Optimization and Bounded Constraints

16 questions

Question 1Question

A small workshop produces two items, Item X and Item Y. Each unit of Item X requires 22 hours of labor and yields a profit of $30\$30. Each unit of Item Y requires 44 hours of labor and yields a profit of $50\$50. The workshop has a maximum of 1616 labor hours available daily and must produce a combined total of at least 55 units per day. Assuming only whole units can be produced, which combination of (Item X, Item Y) maximizes total daily profit while satisfying all given constraints?

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Answer: 6 units of Item X and 1 unit of Item Y

Answer

6 units of Item X and 1 unit of Item Y
The pair consisting of 6 units of Item X and 1 unit of Item Y uses 2(6)+4(1)=162(6) + 4(1) = 16 labor hours (satisfying the maximum 1616-hour limit) and produces 6+1=76 + 1 = 7 units (satisfying the minimum 55-unit limit). It yields a profit of 30(6)+50(1)=$23030(6) + 50(1) = \$230, which is higher than all other feasible options.

Step-by-Step Solution

1
Formulate the constraint inequalities and profit objective function.
Labor constraint: 2X+4Y162X + 4Y \le 16. Minimum unit constraint: X+Y5X + Y \ge 5, where X,Y0X, Y \ge 0 are integers. Profit function: P=30X+50YP = 30X + 50Y.
Setting up the mathematical relations allows systematic evaluation of candidate pairs.
2
Evaluate feasible integer combinations (X,Y)(X, Y) against both constraints.
Combination (6,1)(6, 1): Labor =1616= 16 \le 16, Units =75= 7 \ge 5. Profit =30(6)+50(1)=$230= 30(6) + 50(1) = \$230.
Combination (4,2)(4, 2): Labor =1616= 16 \le 16, Units =65= 6 \ge 5. Profit =30(4)+50(2)=$220= 30(4) + 50(2) = \$220.
Combination (2,3)(2, 3): Labor =1616= 16 \le 16, Units =55= 5 \ge 5. Profit =30(2)+50(3)=$210= 30(2) + 50(3) = \$210.
Evaluating all valid pairs determines which produces the peak profit value.
3
Compare profit outcomes across feasible pairs.
The combination of 66 units of Item X and 11 unit of Item Y achieves the highest valid profit of $230\$230.
The option (6,1)(6, 1) optimizes the objective function under all bounded constraints.

Key Concept

Optimization under linear integer and inequality bounds
Question 2Question

A factory manufactures two products, Product X and Product Y. Daily production is subject to the following constraints:
- Each unit of Product X requires 11 hour of machine time.
- Each unit of Product Y requires 22 hours of machine time.
- Total machine time available per day is at most 1010 hours.
- The factory must produce at least 22 units of Product X and at least 22 units of Product Y per day.
- Each unit of Product X yields a profit of $15\$15, and each unit of Product Y yields a profit of $20\$20.

Match each optimization metric on the left with its corresponding correct numerical value on the right.

Click a left item, then click its matching right item

Items

Maximum possible production of Product Y (in units)
Number of units of Product X required to maximize total daily profit
Maximum total daily profit achievable (in dollars)

Matches

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Answer

Maximum possible production of Product Y matches 4; Units of Product X for maximum profit matches 6; Maximum total daily profit matches 130.
For the maximum production of Product Y, using the minimum required x=2x = 2 allows 2y=102=82y = 10 - 2 = 8, giving y=4y = 4. For maximum profit, each machine hour used on Product X yields $15\$15 per hour while each hour used on Product Y yields $20/2=$10\$20 / 2 = \$10 per hour. Thus, maximizing hours allocated to Product X subject to y2y \ge 2 gives y=2y = 2 and x=6x = 6, yielding a maximum profit of 15(6)+20(2)=13015(6) + 20(2) = 130 dollars.

Step-by-Step Solution

1
Formulate the system of constraints and profit objective function from the narrative context.
Let xx be units of Product X and yy be units of Product Y. Constraints are x2x \ge 2, y2y \ge 2, and x+2y10x + 2y \le 10. Objective is to maximize P=15x+20yP = 15x + 20y.
Establishing explicit algebraic inequalities defines the feasible decision region.
2
Calculate the upper bound for yy.
Setting x=2x = 2 (its minimum value), 2+2y10    2y8    y42 + 2y \le 10 \implies 2y \le 8 \implies y \le 4. Maximum y=4y = 4.
Minimizing xx leaves maximum machine time capacity available for Product Y.
3
Evaluate candidate feasible integer solutions (x,y)(x, y) to determine maximum daily profit.
Valid integer pairs on the boundary include (2,4)(2, 4) with P=110P = 110; (4,3)(4, 3) with P=120P = 120; and (6,2)(6, 2) with P=130P = 130.
Trading 1 unit of Product Y (losing $20\$20) for 2 units of Product X (gaining 2×$15=$302 \times \$15 = \$30) increases net profit by $10\$10 per unit trade.
4
Match each requested metric to its evaluated value.
Maximum Product Y = 4; Product X for max profit = 6; Maximum profit = 130.
Matches each question prompt to its calculated mathematical result.

Key Concept

Linear optimization under inequality bounds and integer constraints.
Question 3Question

A logistics company operates two types of delivery vehicles: Type X and Type Y.

- Each Type X vehicle carries 4040 packages and costs $300\$300 per trip.
- Each Type Y vehicle carries 2525 packages and costs $150\$150 per trip.

The company operates under a maximum daily budget of $3,300\$3,300. Additionally, due to fleet maintenance regulations, the number of Type Y vehicle trips cannot exceed twice the number of Type X vehicle trips.

Which combination of Type X and Type Y vehicle trips maximizes the total package delivery capacity while satisfying all daily budget and maintenance constraints?

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Answer: 6 Type X vehicles and 10 Type Y vehicles

Answer

6 Type X vehicles and 10 Type Y vehicles
The combination of 6 Type X vehicles and 10 Type Y vehicles strictly satisfies the budget constraint (300×6+150×10=33003300300 \times 6 + 150 \times 10 = 3300 \le 3300) and the maintenance constraint (102×6=1210 \le 2 \times 6 = 12). It produces a maximum valid package capacity of 490.

Step-by-Step Solution

1
Formulate the algebraic constraint inequalities
Budget constraint: 300X+150Y3300    2X+Y22300X + 150Y \le 3300 \implies 2X + Y \le 22. Maintenance constraint: Y2XY \le 2X.
Linear constraints define the feasible region for vehicle counts.
2
Express total capacity CC in terms of XX along the budget boundary
C=40X+25YC = 40X + 25Y. On the boundary Y=222XY = 22 - 2X, C=40X+25(222X)=55010XC = 40X + 25(22 - 2X) = 550 - 10X.
To maximize capacity, smaller values of XX along the boundary yield higher total capacity because Type Y provides more capacity per dollar.
3
Find the intersection of constraints to locate the optimal integer point
Setting Y=2XY = 2X into 2X+Y=222X + Y = 22 gives 4X=22    X=5.54X = 22 \implies X = 5.5. The smallest valid integer X5.5X \ge 5.5 satisfying Y=222X2XY = 22 - 2X \le 2X is X=6X = 6.
When X=6X = 6, Y=222(6)=10Y = 22 - 2(6) = 10. Checking maintenance: 102(6)=1210 \le 2(6) = 12 holds true.
4
Calculate the maximum capacity
C=40(6)+25(10)=240+250=490C = 40(6) + 25(10) = 240 + 250 = 490 packages.
This point (6,10)(6, 10) yields the maximum capacity satisfying all linear inequality constraints.

Key Concept

Linear Optimization under Bounded Constraints
Question 4Question

An analytics firm processes data batches using two types of cloud computing instances: High-Memory Instances (MM) and High-Compute Instances (CC).

- Processing Capacity: Each High-Memory Instance processes 120120 thousand transactions per hour, and each High-Compute Instance processes 200200 thousand transactions per hour. The workload requires a total processing rate of at least 1,3201,320 thousand transactions per hour.
- Instance Availability: At most 1010 High-Memory Instances (M10M \le 10) and at most 66 High-Compute Instances (C6C \le 6) are available. At least 11 instance of each type must be used (M1M \ge 1 and C1C \ge 1).
- Load Balancing Constraint: To ensure infrastructure stability, the number of High-Memory Instances cannot exceed twice the number of High-Compute Instances (M2CM \le 2C).
- Operating Costs: High-Memory Instances cost $18\$18 per hour each, while High-Compute Instances cost $25\$25 per hour each.

Select the number of High-Memory Instances (MM) and the number of High-Compute Instances (CC) that minimize the total hourly operating cost while satisfying all operational requirements.

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Items

Number of High-Memory Instances (MM)
Number of High-Compute Instances (CC)

Matches

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Answer

The minimum cost is achieved by selecting 1 High-Memory Instance (M=1M = 1) and 6 High-Compute Instances (C=6C = 6).
To minimize the total hourly cost subject to 3M+5C333M + 5C \ge 33, M2CM \le 2C, 1M101 \le M \le 10, and 1C61 \le C \le 6, testing all feasible integer boundary points shows that C=6C = 6 allows M=1M = 1 (since 3(1)+5(6)=333(1) + 5(6) = 33 and 12(6)1 \le 2(6) holds), producing the lowest cost of $168\$168. Thus, M=1M = 1 and C=6C = 6 are the optimal selections.

Step-by-Step Solution

1
Set up the objective function and mathematical constraints from the problem stem.
Objective: Minimize Cost=18M+25C\text{Cost} = 18M + 25C.
Constraints:
1) Throughput: 120M+200C1320    3M+5C33120M + 200C \ge 1320 \implies 3M + 5C \ge 33.
2) Availability: 1M101 \le M \le 10 and 1C61 \le C \le 6 (integers).
3) Load balancing: M2CM \le 2C.
Formulating the inequalities defines the boundary of feasible integer pairs (M,C)(M, C).
2
Test candidate integer boundary values for C{1,2,3,4,5,6}C \in \{1, 2, 3, 4, 5, 6\}.
- For C=1,2C = 1, 2: Maximum M2CM \le 2C yields 3(2C)+5C=11C3(2C) + 5C = 11C, giving max 1111 (for C=1C=1) and 2222 (for C=2C=2), both <33< 33 (infeasible).
- For C=3C = 3: M6M \le 6. 3M335(3)=18    M63M \ge 33 - 5(3) = 18 \implies M \ge 6. Min M=6M = 6. Cost =18(6)+25(3)=108+75=183= 18(6) + 25(3) = 108 + 75 = 183.
- For C=4C = 4: M8M \le 8. 3M335(4)=13    M53M \ge 33 - 5(4) = 13 \implies M \ge 5. Min M=5M = 5. Cost =18(5)+25(4)=90+100=190= 18(5) + 25(4) = 90 + 100 = 190.
- For C=5C = 5: M10M \le 10. 3M335(5)=8    M33M \ge 33 - 5(5) = 8 \implies M \ge 3. Min M=3M = 3. Cost =18(3)+25(5)=54+125=179= 18(3) + 25(5) = 54 + 125 = 179.
- For C=6C = 6: M10M \le 10. 3M335(6)=3    M13M \ge 33 - 5(6) = 3 \implies M \ge 1. Min M=1M = 1. Cost =18(1)+25(6)=18+150=168= 18(1) + 25(6) = 18 + 150 = 168.
Evaluating minimal feasible MM for each allowed CC identifies all vertex and boundary candidates.
3
Compare total cost across all valid candidate combinations.
The candidate pair (M=1,C=6)(M=1, C=6) yields the minimum operating cost of $168\$168, while satisfying 120(1)+200(6)=13201320120(1) + 200(6) = 1320 \ge 1320 and 12(6)1 \le 2(6).
Comparing all valid boundary points proves that M=1M=1 and C=6C=6 minimize cost under all constraints.

Key Concept

Optimization under Bounded Constraints
Question 5Question

A commercial greenhouse allocates integer numbers of acres to grow two high-yield crops: Organic Tomatoes (TT) and Hydroponic Cucumbers (CC). The operational parameters and constraints are as follows:

- Each acre of Tomatoes requires 33 units of water per day and 44 units of fertilizer per week, producing a net revenue of $1,200\$1,200 per week.
- Each acre of Cucumbers requires 55 units of water per day and 22 units of fertilizer per week, producing a net revenue of $1,000\$1,000 per week.
- Total daily water usage across both crops cannot exceed 4545 units.
- Total weekly fertilizer usage across both crops cannot exceed 3636 units.
- To fulfill distributor agreements, the total cultivated area (T+CT + C) must be at least 1010 acres.

Which of the following crop allocation pairs (T,C)(T, C) maximizes total weekly net revenue while satisfying all operational constraints?

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Answer: 77 acres of Tomatoes and 44 acres of Cucumbers

Answer

The optimal crop allocation is 7 acres of Tomatoes and 4 acres of Cucumbers, yielding a maximum weekly revenue of $12,400.
The allocation of 7 acres of Tomatoes and 4 acres of Cucumbers fully respects all three operational boundaries: daily water (41 out of 45 units), weekly fertilizer (36 out of 36 units), and total cultivated area (11 acres, meeting the minimum threshold of 10 acres). It produces the highest valid weekly revenue of $12,400.

Step-by-Step Solution

1
Formulate the mathematical model from the context
Maximize Revenue R=1200T+1000CR = 1200T + 1000C subject to:
1) Water constraint: 3T+5C453T + 5C \le 45
2) Fertilizer constraint: 4T+2C36    2T+C184T + 2C \le 36 \implies 2T + C \le 18
3) Land constraint: T+C10T + C \ge 10
4) Integer constraints: T,CZ0T, C \in \mathbb{Z}_{\ge 0}
Establishing explicit linear inequality bounds and the objective function is required for constrained optimization.
2
Evaluate feasible integer boundary points near the intersection of joint constraints
Testing boundary candidate pairs:
- At T=7,C=4T = 7, C = 4: Water =3(7)+5(4)=4145= 3(7)+5(4) = 41 \le 45, Fertilizer =4(7)+2(4)=3636= 4(7)+2(4) = 36 \le 36, Area =1110= 11 \ge 10. Feasible! Revenue =1200(7)+1000(4)=$12,400= 1200(7) + 1000(4) = \$12,400.
- At T=8,C=2T = 8, C = 2: Water =3445= 34 \le 45, Fertilizer =3636= 36 \le 36, Area =1010= 10 \ge 10. Feasible! Revenue =1200(8)+1000(2)=$11,600= 1200(8) + 1000(2) = \$11,600.
- At T=6,C=5T = 6, C = 5: Water =4345= 43 \le 45, Fertilizer =3436= 34 \le 36, Area =1110= 11 \ge 10. Feasible! Revenue =1200(6)+1000(5)=$12,200= 1200(6) + 1000(5) = \$12,200.
Systematic evaluation of feasible integer boundary points identifies the combination that maximizes net revenue.
3
Verify that higher revenue candidates violate joint system bounds
Checking candidate (8,3)(8, 3): Revenue would be $12,600\$12,600, but Fertilizer =4(8)+2(3)=38>36= 4(8) + 2(3) = 38 > 36 (infeasible).
Ensures that candidates yielding higher revenue are invalid due to joint constraint violations.

Key Concept

Optimization under Bounded Linear Constraints
Estimated Time:2m 30s
Question 6Question

A research laboratory is formulating a daily dosage protocol for a clinical trial combining two therapeutics, Drug A (AA) and Drug B (BB), measured in integer milligrams (mg).

The trial protocol specifies the following operational constraints:
- The daily dosage of Drug A must be at least 10 mg10\text{ mg} and at most 40 mg40\text{ mg} (10A4010 \leq A \leq 40).
- The daily dosage of Drug B must be at least 15 mg15\text{ mg} and at most 50 mg50\text{ mg} (15B5015 \leq B \leq 50).
- To prevent hepatotoxicity, the combined daily dosage (A+B)(A + B) cannot exceed 65 mg65\text{ mg} (A+B65A + B \leq 65).
- To ensure therapeutic efficacy, the dosage of Drug B must be at least 10 mg10\text{ mg} less than twice the dosage of Drug A (B2A10B \geq 2A - 10).

The total treatment efficacy score EE is modeled by the linear function E=3A+4BE = 3A + 4B.

Based on the constraints above, select the daily dosage for Drug A and the daily dosage for Drug B that together maximize the total treatment efficacy score EE.

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Items

Daily Dosage of Drug A (mg)
Daily Dosage of Drug B (mg)

Matches

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Answer

The optimal daily dosage is 15 mg for Drug A and 50 mg for Drug B, yielding a maximum total efficacy score of 245.
Because Drug B contributes +4+4 points per milligram to efficacy while Drug A contributes +3+3 points per milligram, maximizing Drug B to its upper limit of 50 mg provides the largest gain in total score. Given B=50 mgB = 50\text{ mg}, the combined toxicity limit A+B65 mgA + B \leq 65\text{ mg} restricts Drug A to at most 15 mg15\text{ mg}. Testing (A,B)=(15,50)(A, B) = (15, 50) satisfies all individual and joint constraints (151015 \geq 10, 502(15)10=2050 \geq 2(15) - 10 = 20), yielding the maximum overall score of 245245.

Step-by-Step Solution

1
Identify the objective function and system of linear inequalities.
Objective: Maximize E=3A+4BE = 3A + 4B subject to 10A4010 \leq A \leq 40, 15B5015 \leq B \leq 50, A+B65A + B \leq 65, and 2AB102A - B \leq 10.
Establishing the mathematical model allows systematically evaluating boundary vertices.
2
Analyze the objective function weights to determine optimization direction.
Drug B has a higher coefficient (+4+4) than Drug A (+3+3).
Increasing BB yields a greater increase in total efficacy EE per milligram than increasing AA, so BB should be made as large as allowed by constraints.
3
Test the maximum individual bound for Drug B.
Setting B=50B = 50 (its maximum bound), the toxicity constraint becomes A+5065    A15A + 50 \leq 65 \implies A \leq 15.
Determining the largest feasible AA given the maximal value of BB maximizes 3A+4(50)3A + 4(50).
4
Verify all secondary constraints at candidate point (A,B)=(15,50)(A, B) = (15, 50).
1) 10154010 \leq 15 \leq 40 (Satisfied)
2) 15505015 \leq 50 \leq 50 (Satisfied)
3) 15+50=656515 + 50 = 65 \leq 65 (Satisfied)
4) 502(15)10=2050 \geq 2(15) - 10 = 20 (Satisfied)
Total Efficacy: E=3(15)+4(50)=45+200=245E = 3(15) + 4(50) = 45 + 200 = 245.
Ensures the corner candidate point is fully compliant with every given restriction.
5
Compare against alternative valid vertex candidates.
Candidate (10,50)    E=30+200=230(10, 50) \implies E = 30 + 200 = 230.
Candidate (25,40)    E=75+160=235(25, 40) \implies E = 75 + 160 = 235.
Candidate (40,25)    E=120+100=220(40, 25) \implies E = 120 + 100 = 220.
Confirms that (15,50)(15, 50) yields the maximum value among all feasible integer combinations.

Key Concept

Optimization under Linear Inequality Bounded Constraints
Question 7Question

A pharmaceutical laboratory synthesizes two custom therapeutic compounds, Compound XX and Compound YY, in integer batch quantities during a single production cycle. Production is subject to the following resource and operational constraints:

- Bioreactor Time: Each batch of Compound XX requires 44 hours and each batch of Compound YY requires 33 hours. The laboratory has at most 4848 total hours of bioreactor time available.
- Specialized Solvent: Each batch of Compound XX requires 33 liters and each batch of Compound YY requires 55 liters. The total solvent supply is capped at 5555 liters.
- Catalyst Stability Limit: To prevent reactive degradation, the number of batches of Compound YY produced cannot exceed twice the number of batches of Compound XX plus 11 (Y2X+1Y \le 2X + 1).

Each batch of Compound XX generates a net profit of $12,000\$12,000, and each batch of Compound YY generates a net profit of $14,000\$14,000.

What is the maximum total net profit, in dollars, that the laboratory can achieve within these combined production constraints?

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Answer: $172,000\$172,000

Answer

The maximum total net profit achievable under all constraints is $172,000\$172,000, obtained by producing 55 batches of Compound XX and 88 batches of Compound YY.
The production plan of 55 batches of Compound XX and 88 batches of Compound YY satisfies all given constraints: Bioreactor time (4(5)+3(8)=44484(5) + 3(8) = 44 \le 48 hours), Solvent volume (3(5)+5(8)=55553(5) + 5(8) = 55 \le 55 liters), and Catalyst stability (82(5)+1=118 \le 2(5) + 1 = 11). This combination yields the maximum valid profit of 5×$12,000+8×$14,000=$172,0005 \times \$12,000 + 8 \times \$14,000 = \$172,000.

Step-by-Step Solution

1
Formulate the linear inequalities representing the system constraints.
Bioreactor constraint: 4X+3Y484X + 3Y \le 48; Solvent constraint: 3X+5Y553X + 5Y \le 55; Catalyst constraint: Y2X+1Y \le 2X + 1; Non-negativity and integer constraints: X,YZ0X, Y \in \mathbb{Z}_{\ge 0}. Objective function: Maximize P=12000X+14000YP = 12000X + 14000Y.
Establishing the mathematical boundaries of the feasible region for integer optimization.
2
Find the intersection point of the two primary resource constraint boundaries (4X+3Y=484X + 3Y = 48 and 3X+5Y=553X + 5Y = 55).
Multiplying 4X+3Y=484X + 3Y = 48 by 55 yields 20X+15Y=24020X + 15Y = 240. Multiplying 3X+5Y=553X + 5Y = 55 by 33 yields 9X+15Y=1659X + 15Y = 165. Subtracting the two equations gives 11X=75    X=75/116.8211X = 75 \implies X = 75/11 \approx 6.82, leading to Y=76/116.91Y = 76/11 \approx 6.91.
Identifies the continuous optimal vertex of the system to narrow down nearby integer search points.
3
Test nearby integer pairs (X,Y)(X, Y) within the feasible region.
For X=6X = 6, max YY under solvent limit is (5518)/5=7\lfloor(55 - 18)/5\rfloor = 7. Check (6,7)(6, 7): 4(6)+3(7)=45484(6)+3(7)=45 \le 48, 3(6)+5(7)=53553(6)+5(7)=53 \le 55, 72(6)+1=137 \le 2(6)+1=13. Profit: 12000(6)+14000(7)=$170,00012000(6)+14000(7) = \$170,000.
For X=5X = 5, max YY under solvent limit is (5515)/5=8\lfloor(55 - 15)/5\rfloor = 8. Check (5,8)(5, 8): 4(5)+3(8)=44484(5)+3(8)=44 \le 48, 3(5)+5(8)=55553(5)+5(8)=55 \le 55, 82(5)+1=118 \le 2(5)+1=11. Profit: 12000(5)+14000(8)=$172,00012000(5)+14000(8) = \$172,000.
For X=4X = 4, max YY under solvent limit is (5512)/5=8\lfloor(55 - 12)/5\rfloor = 8. Profit: 12000(4)+14000(8)=$160,00012000(4)+14000(8) = \$160,000.
For X=7X = 7, max YY under bioreactor limit is (4828)/3=6\lfloor(48 - 28)/3\rfloor = 6. Check (7,6)(7, 6): 4(7)+3(6)=46484(7)+3(6)=46 \le 48, 3(7)+5(6)=51553(7)+5(6)=51 \le 55. Profit: 12000(7)+14000(6)=$168,00012000(7)+14000(6) = \$168,000.
Evaluates candidate lattice points on the boundary of the feasible integer region.
4
Compare profit totals among all valid integer combinations.
The combination (5,8)(5, 8) yields the maximum valid profit of $172,000\$172,000.
Confirms the global maximum integer solution under all joint bounded constraints.

Key Concept

Optimization under Integer and Joint Bounded Constraints
Question 8Question

An earth-observation satellite payload operator allocates integer numbers of channels to two operational payload modes: High-Resolution Imaging (II) and Atmospheric Sounding (SS). Operational parameters and resource bounds are specified as follows:

- Bandwidth: Each imaging channel requires 15 MHz15\text{ MHz} and each sounding channel requires 10 MHz10\text{ MHz}. Total available payload bandwidth is at most 210 MHz210\text{ MHz}.
- Power: Each imaging channel consumes 40 W40\text{ W} and each sounding channel consumes 15 W15\text{ W}. Total available power budget is at most 440 W440\text{ W}.
- Mission Minimums: The satellite must operate at least 22 imaging channels (I2I \ge 2) and at least 44 sounding channels (S4S \ge 4).
- Buffer Constraint: The number of sounding channels cannot exceed twice the number of imaging channels (S2IS \le 2I).

Match each payload optimization target on the left with its corresponding integer channel count on the right.

Click a left item, then click its matching right item

Items

Maximum possible number of Imaging channels (II)
Maximum possible number of Sounding channels (SS)
Number of Sounding channels (SS) that maximizes daily data return (R=70I+20SR = 70I + 20S Gbit/day)

Matches

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Answer

The maximum possible number of Imaging channels (II) is 9. The maximum possible number of Sounding channels (SS) is 12. The number of Sounding channels (SS) that maximizes daily data return is 5.
Matching the maximum imaging channels (II) to 9 is correct because I=10I = 10 forces S2S \le 2, violating S4S \ge 4, whereas I=9I = 9 is fully feasible with S=5S = 5. Matching the maximum sounding channels (SS) to 12 is correct because at (6,12)(6, 12), bandwidth, power, and buffer constraints are simultaneously satisfied. Matching the sounding channels for maximum data return to 5 is correct because the function R=70I+20SR = 70I + 20S reaches its peak of 730 Gbit/day at (9,5)(9, 5).

Step-by-Step Solution

1
Formulate the algebraic system of inequality constraints.
Bandwidth: 15I+10S210    3I+2S4215I + 10S \le 210 \implies 3I + 2S \le 42. Power: 40I+15S440    8I+3S8840I + 15S \le 440 \implies 8I + 3S \le 88. Bounds: I2I \ge 2, S4S \ge 4, and S2IS \le 2I.
Converting narrative bounds into simplified algebraic inequalities facilitates evaluating integer extreme points.
2
Find the maximum feasible value for Imaging channels (II).
If I=10I = 10, power requires 8(10)+3S88    3S8    S28(10) + 3S \le 88 \implies 3S \le 8 \implies S \le 2, which violates the minimum requirement S4S \ge 4. If I=9I = 9, power requires 8(9)+3S88    3S16    S58(9) + 3S \le 88 \implies 3S \le 16 \implies S \le 5. With S=4S = 4 or S=5S = 5, bandwidth gives 3(9)+2(5)=37423(9) + 2(5) = 37 \le 42 and S2(9)S \le 2(9) holds. Thus, max I=9I = 9.
Determining the upper bound of II requires testing larger integer values of II while ensuring S4S \ge 4 is not violated.
3
Find the maximum feasible value for Sounding channels (SS).
Since S2IS \le 2I, SS is constrained by II. Testing I=6I = 6: bandwidth gives 3(6)+2S42    2S24    S123(6) + 2S \le 42 \implies 2S \le 24 \implies S \le 12. Power gives 8(6)+3S88    3S40    S13.338(6) + 3S \le 88 \implies 3S \le 40 \implies S \le 13.33. Buffer gives S2(6)=12S \le 2(6) = 12. At (6,12)(6, 12), all constraints hold (18+24=424218+24=42 \le 42, 48+36=848848+36=84 \le 88, 121212 \le 12). Thus, max S=12S = 12.
Maximizing SS requires balancing the ratio constraint S2IS \le 2I against bandwidth and power limits.
4
Maximize the objective function R(I,S)=70I+20SR(I, S) = 70I + 20S over all feasible boundary points.
Evaluating candidate integer vertex points: R(9,5)=70(9)+20(5)=730R(9, 5) = 70(9) + 20(5) = 730; R(9,4)=710R(9, 4) = 710; R(8,8)=720R(8, 8) = 720; R(7,10)=690R(7, 10) = 690; R(6,12)=660R(6, 12) = 660. The maximum data return occurs at (9,5)(9, 5), where S=5S = 5.
Comparing total return at all boundary points identifies the global integer maximum.

Key Concept

Linear Programming and Integer Bounded Optimization
Question 9Question

A regional logistics operator manages cargo transit on a river corridor using two types of vessels: Express Barges (EE) and Heavy-Haul Barges (HH). Weekly operational constraints are defined as follows:

- Fuel Allowance: Each Express Barge consumes 88 metric tons of fuel per trip, and each Heavy-Haul Barge consumes 1212 metric tons. Total weekly fuel consumption cannot exceed 144144 metric tons.
- Crew Availability: Each Express Barge requires 33 crew shifts, and each Heavy-Haul Barge requires 22 crew shifts. Total available crew shifts per week cannot exceed 3030.
- Service Minimum: The operator must deploy at least 22 Heavy-Haul Barges (H2H \ge 2) per week to maintain baseline commercial obligations.
- Profit Structure: Each Express Barge generates a net profit of $5,000\$5,000, and each Heavy-Haul Barge generates a net profit of $6,000\$6,000.

Match each operational metric on the left with its correct optimal value under the profit-maximizing schedule of integer barge deployments.

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Items

Number of Express Barges (EE) in the profit-maximizing schedule
Number of Heavy-Haul Barges (HH) in the profit-maximizing schedule
Maximum total weekly profit achieved (in thousands of dollars)

Matches

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Answer

The profit-maximizing deployment requires 33 Express Barges and 1010 Heavy-Haul Barges, yielding a total maximum profit of 7575 thousand dollars ($75,000\$75,000).
The correct matches establish that deploying 33 Express Barges and 1010 Heavy-Haul Barges yields the maximum possible weekly profit of $75,000\$75,000. This combination uses 2(3)+3(10)=362(3) + 3(10) = 36 tons of fuel (using 100%100\% of fuel capacity) and 3(3)+2(10)=293(3) + 2(10) = 29 crew shifts (out of 3030 available), satisfying all joint constraints.

Step-by-Step Solution

1
Formulate the linear programming inequalities from the narrative constraints.
Fuel constraint: 8E+12H144    2E+3H368E + 12H \le 144 \implies 2E + 3H \le 36.
Crew constraint: 3E+2H303E + 2H \le 30.
Bounds: H2H \ge 2, E0E \ge 0, with E,HZE, H \in \mathbb{Z}.
Profit function: P(E,H)=5E+6HP(E, H) = 5E + 6H (in thousands of dollars).
Converting real-world bounds into simplified algebraic inequalities is required to determine the feasible integer region.
2
Identify boundary integer candidate points near the intersection of the constraints.
Solving 2E+3H=362E + 3H = 36 and 3E+2H=303E + 2H = 30 gives 5H=48    H=9.6,E=3.65H = 48 \implies H = 9.6, E = 3.6.
Testing adjacent integer points (E,H)(E, H) within the feasible region:
- Case 1: H=10    2E+3036    E3H = 10 \implies 2E + 30 \le 36 \implies E \le 3. Check crew for (3,10)(3, 10): 3(3)+2(10)=29303(3) + 2(10) = 29 \le 30 (Feasible). Profit P=5(3)+6(10)=75P = 5(3) + 6(10) = 75.
- Case 2: H=9    2E+2736    E4H = 9 \implies 2E + 27 \le 36 \implies E \le 4. Check crew for (4,9)(4, 9): 3(4)+2(9)=30303(4) + 2(9) = 30 \le 30 (Feasible). Profit P=5(4)+6(9)=74P = 5(4) + 6(9) = 74.
- Case 3: H=12    2E+3636    E=0H = 12 \implies 2E + 36 \le 36 \implies E = 0. Check crew for (0,12)(0, 12): 3(0)+2(12)=24303(0) + 2(12) = 24 \le 30 (Feasible). Profit P=6(12)=72P = 6(12) = 72.
- Case 4: H=11    2E+3336    E1H = 11 \implies 2E + 33 \le 36 \implies E \le 1. Check crew for (1,11)(1, 11): 3(1)+2(11)=25303(1) + 2(11) = 25 \le 30 (Feasible). Profit P=5(1)+6(11)=71P = 5(1) + 6(11) = 71.
Since integer constraints apply, evaluating feasible grid points along the constraint boundary identifies the global maximum.
3
Compare profit outcomes and confirm optimal values.
The maximum profit occurs at (E,H)=(3,10)(E, H) = (3, 10), giving a profit of $75,000\$75,000.
Comparing all candidate integer points verifies that E=3E=3, H=10H=10, and P=75P=75 is the unique optimal solution.

Key Concept

Integer Optimization under Simultaneous Bounded Constraints
Question 10Question

A cloud analytics company configures daily data processing operations using two types of virtual server instances: Compute-Optimized (CC) and Memory-Optimized (MM). The operational parameters and constraints are as follows:

- Each CC instance processes 50 batch jobs per hour and incurs an operating cost of 15perhour.Each15 per hour. - Each M instanceprocesses30batchjobsperhourandincursanoperatingcostof instance processes 30 batch jobs per hour and incurs an operating cost of 10 per hour.
- The system must process at least 360 batch jobs per hour in total.
- The total hourly operating budget for server instances cannot exceed 115.Duetoactivedatabasecachingpolicies,atleast3115. - Due to active database caching policies, at least 3 M instancesmustbeactiveatalltimes( instances must be active at all times ( M \ge 3 ).). - C and and M$ must both be non-negative integers.

Which of the following combinations of Compute (CC) and Memory (MM) server instances maximizes the total hourly job throughput while satisfying all operational constraints?

Show answer & explanation

Answer: 5 Compute instances and 4 Memory instances

Answer

5 Compute instances and 4 Memory instances
The combination of 5 Compute instances and 4 Memory instances satisfies all stated system bounds (15×5+10×4=11511515 \times 5 + 10 \times 4 = 115 \le 115, 50×5+30×4=37036050 \times 5 + 30 \times 4 = 370 \ge 360, and 434 \ge 3) and delivers the maximum feasible throughput of 370 jobs per hour.

Step-by-Step Solution

1
Formulate the linear system of inequalities from the given constraints.
Throughput: 50C+30M3605C+3M3650C + 30M \ge 360 \Rightarrow 5C + 3M \ge 36.
Budget: 15C+10M1153C+2M2315C + 10M \le 115 \Rightarrow 3C + 2M \le 23.
Minimum MM: M3M \ge 3, where C,MZ0C, M \in \mathbb{Z}_{\ge 0}.
Converting narrative conditions into simplified algebraic inequalities makes constraint checking straightforward.
2
Determine the set of integer pairs (C,M)(C, M) that satisfy all three inequalities.
Combining 3C+2M233C + 2M \le 23 and 5C+3M365C + 3M \ge 36 with M3M \ge 3 yields only two valid integer pairs:
- Pair 1: (C=3,M=7)(C=3, M=7) with cost 15(3)+10(7)=11515(3)+10(7)=115 and throughput 50(3)+30(7)=36050(3)+30(7)=360
- Pair 2: (C=5,M=4)(C=5, M=4) with cost 15(5)+10(4)=11515(5)+10(4)=115 and throughput 50(5)+30(4)=37050(5)+30(4)=370
Systematic testing of integer values bounded by the linear inequalities identifies all feasible solutions.
3
Compare total throughput across feasible solutions to find the maximum.
Pair 2 (C=5,M=4)(C=5, M=4) produces 370 jobs per hour, which exceeds Pair 1's 360 jobs per hour.
The question specifically asks to maximize throughput among all valid configurations.

Key Concept

Optimization under linear integer constraints and bounded systems
Question 11Question

A municipal transit agency is installing two types of electric vehicle (EV) charging stations at a new central station: Level 2 Chargers (LL) and Level 3 Fast Chargers (DD). Each Level 2 charger costs $4,000\$4,000 to install and draws 10 kW10\text{ kW} of grid power. Each Level 3 charger costs $15,000\$15,000 to install and draws 50 kW50\text{ kW} of grid power. The agency has a total installation budget of $100,000\$100,000 and a maximum available grid power allocation of 320 kW320\text{ kW}. To meet minimum service grant requirements, the hub must install at least 5 Level 2 chargers (L5L \ge 5) and at least 3 Level 3 chargers (D3D \ge 3). Each Level 2 charger can serve up to 8 sessions per day, and each Level 3 charger can serve up to 30 sessions per day. Match each operational optimization metric on the left with its corresponding correct value on the right that maximizes the total daily charging sessions served by the hub.

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Items

Optimal number of Level 2 chargers (LL)
Optimal number of Level 3 chargers (DD)
Maximum total daily charging sessions

Matches

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Answer

The optimal operational configuration consists of 10 Level 2 chargers and 4 Level 3 chargers, achieving a maximum total of 200 daily charging sessions.
To maximize daily sessions S=8L+30DS = 8L + 30D under budget 4L+15D1004L + 15D \le 100, power L+5D32L + 5D \le 32, and lower bounds L5,D3L \ge 5, D \ge 3, testing valid integer points shows that setting Level 3 chargers to 4 and Level 2 chargers to 10 satisfies all joint constraints while achieving the peak capacity of 200 total daily sessions.

Step-by-Step Solution

1
Formulate the linear system of constraints and objective function.
Budget constraint: 4,000L+15,000D100,000    4L+15D1004,000L + 15,000D \le 100,000 \implies 4L + 15D \le 100.
Power constraint: 10L+50D320    L+5D3210L + 50D \le 320 \implies L + 5D \le 32.
Minimum bounds: L5L \ge 5 and D3D \ge 3.
Objective function to maximize: S=8L+30DS = 8L + 30D.
Define the mathematical boundaries and optimization goal based on problem parameters.
2
Evaluate feasible integer values for Level 3 chargers (DD).
Case 1: If D=3D = 3:
- Budget bound: 4L+15(3)100    4L55    L134L + 15(3) \le 100 \implies 4L \le 55 \implies L \le 13.
- Power bound: L+5(3)32    L17L + 5(3) \le 32 \implies L \le 17.
- Max feasible L=13L = 13.
- Total sessions S=8(13)+30(3)=104+90=194S = 8(13) + 30(3) = 104 + 90 = 194.

Case 2: If D=4D = 4:
- Budget bound: 4L+15(4)100    4L40    L104L + 15(4) \le 100 \implies 4L \le 40 \implies L \le 10.
- Power bound: L+5(4)32    L12L + 5(4) \le 32 \implies L \le 12.
- Max feasible L=10L = 10.
- Total sessions S=8(10)+30(4)=80+120=200S = 8(10) + 30(4) = 80 + 120 = 200.

Case 3: If D=5D = 5:
- Budget bound: 4L+15(5)100    4L25    L64L + 15(5) \le 100 \implies 4L \le 25 \implies L \le 6.
- Power bound: L+5(5)32    L7L + 5(5) \le 32 \implies L \le 7.
- Max feasible L=6L = 6.
- Total sessions S=8(6)+30(5)=48+150=198S = 8(6) + 30(5) = 48 + 150 = 198.

Case 4: If D=6D = 6:
- Budget bound: 4L+15(6)100    4L10    L24L + 15(6) \le 100 \implies 4L \le 10 \implies L \le 2.
- Violates minimum constraint L5L \ge 5.
Systematically check integer combinations of variables within the feasible region to identify maximum daily sessions.
3
Compare total sessions across valid integer pairs.
The maximum session count occurs when D=4D = 4 and L=10L = 10, producing S=200S = 200 daily sessions.
Determine the unique optimal combination satisfying all joint constraints.

Key Concept

Linear Optimization under Bounded Integer Constraints
Estimated Time:2m 0s
Question 12Question

A boutique catering kitchen prepares custom corporate event packages containing two types of dessert selections: Specialty Layer Cakes (LL) and Miniature Cupcake Sets (MM). Each Specialty Layer Cake requires 33 hours of decorating labor and 44 kg of specialized flour. Each Miniature Cupcake Set requires 11 hour of decorating labor and 22 kg of specialized flour. For an upcoming event order, the kitchen has a maximum resource availability of 3030 total hours of decorating labor and 4444 kg of specialized flour. To meet client specifications, the kitchen must produce at least 44 Specialty Layer Cakes. If each Specialty Layer Cake yields a profit of $50\$50 and each Miniature Cupcake Set yields a profit of $20\$20, which of the following pairs of (L,M)(L, M) represents the combination of Specialty Layer Cakes and Miniature Cupcake Sets that maximizes total profit while satisfying all resource constraints?

Show answer & explanation

Answer: (8,6)(8, 6)

Answer

The optimal production combination is 8 Specialty Layer Cakes and 6 Miniature Cupcake Sets, yielding a maximum profit of $520.
The pair (8,6)(8, 6) satisfies the minimum requirement of at least 4 Specialty Layer Cakes (848 \ge 4), requires 3(8)+1(6)=303(8) + 1(6) = 30 labor hours (within the 30-hour limit), and requires 4(8)+2(6)=444(8) + 2(6) = 44 kg of flour (within the 44-kg limit). It yields a total profit of 50(8)+20(6)=$52050(8) + 20(6) = \$520, which is the highest achievable profit among all feasible combinations.

Step-by-Step Solution

1
Formulate the mathematical model for constraints and objective function.
Decorating Labor: 3L+M303L + M \le 30; Specialized Flour: 4L+2M444L + 2M \le 44 (or 2L+M222L + M \le 22); Minimum Requirement: L4L \ge 4; Profit: P=50L+20MP = 50L + 20M.
Establishing explicit linear inequality bounds allows systematic evaluation of feasible integer pairs (L,M)(L, M).
2
Determine upper bound for MM as a function of integer values of L4L \ge 4.
For L=4L=4: Mmin(3012,228)=14    P=50(4)+20(14)=$480M \le \min(30-12, 22-8) = 14 \implies P = 50(4)+20(14) = \$480.
For L=5L=5: Mmin(3015,2210)=12    P=50(5)+20(12)=$490M \le \min(30-15, 22-10) = 12 \implies P = 50(5)+20(12) = \$490.
For L=6L=6: Mmin(3018,2212)=10    P=50(6)+20(10)=$500M \le \min(30-18, 22-12) = 10 \implies P = 50(6)+20(10) = \$500.
For L=7L=7: Mmin(3021,2214)=8    P=50(7)+20(8)=$510M \le \min(30-21, 22-14) = 8 \implies P = 50(7)+20(8) = \$510.
For L=8L=8: Mmin(3024,2216)=6    P=50(8)+20(6)=$520M \le \min(30-24, 22-16) = 6 \implies P = 50(8)+20(6) = \$520.
For L=9L=9: Mmin(3027,2218)=3    P=50(9)+20(3)=$510M \le \min(30-27, 22-18) = 3 \implies P = 50(9)+20(3) = \$510.
For L=10L=10: Mmin(3030,2220)=0    P=50(10)+20(0)=$500M \le \min(30-30, 22-20) = 0 \implies P = 50(10)+20(0) = \$500.
Evaluating potential corner points and boundary integer points identifies the maximum profit within the bounded region.
3
Compare maximum calculated profit across all valid integer pairs.
The highest profit occurs at L=8,M=6L = 8, M = 6 with P=$520P = \$520.
This point satisfies all constraints while maximizing the objective function.

Key Concept

Optimization under multiple joint inequality constraints
Question 13Question

A printing company produces two types of customized marketing materials: Standard brochures (SS) and Premium brochures (PP). Each Standard brochure requires 22 minutes of printing time and 11 minute of binding time. Each Premium brochure requires 33 minutes of printing time and 33 minutes of binding time. The production facility has a maximum daily capacity of 240240 minutes for printing and 180180 minutes for binding. Due to a recurring client agreement, the facility must produce at least 2020 Standard brochures per day. The profit is $4\$4 per Standard brochure and $9\$9 per Premium brochure. Match each optimization variable or outcome on the left with its correct value on the right that maximizes daily total profit while satisfying all operational constraints.

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Items

Optimal number of Standard brochures (SS)
Optimal number of Premium brochures (PP)
Maximum total daily profit (ZZ)

Matches

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Answer

The optimal production values are 6060 Standard brochures and 4040 Premium brochures, resulting in a maximum daily profit of $600\$600.
Solving the system of simultaneous constraints yields the optimal vertex at S=60S = 60 Standard brochures and P=40P = 40 Premium brochures. At this point, both printing (2(60)+3(40)=2402(60) + 3(40) = 240) and binding (60+3(40)=18060 + 3(40) = 180) capacities are fully utilized, achieving the maximum daily profit of 4(60)+9(40)=$6004(60) + 9(40) = \$600.

Step-by-Step Solution

1
Formulate the linear programming model with constraints and objective function.
Constraints: 2S+3P2402S + 3P \le 240 (printing time), S+3P180S + 3P \le 180 (binding time), S20S \ge 20 (contractual minimum), and S,P0S, P \ge 0. Objective: Maximize Z=4S+9PZ = 4S + 9P.
Clear mathematical formulation is required to define the feasible region bounded by resource limits.
2
Calculate the intersection points (vertices) of the boundary lines of the feasible region.
Subtracting S+3P=180S + 3P = 180 from 2S+3P=2402S + 3P = 240 yields S=60S = 60. Substituting S=60S = 60 into S+3P=180S + 3P = 180 gives 3P=120    P=403P = 120 \implies P = 40. Corner point 1: (60,40)(60, 40).
The fundamental theorem of linear programming states that maximum profit occurs at a vertex of the feasible region.
3
Find remaining boundary vertices satisfying all constraints.
Corner point 2: (120,0)(120, 0) from printing limit when P=0P=0 (satisfies binding limit 120180120 \le 180). Corner point 3: (20,53)(20, 53) when S=20S=20 (integer constraint on PP where 20+3P180    P53.3320 + 3P \le 180 \implies P \le 53.33). Corner point 4: (20,0)(20, 0).
All feasible vertices must be identified to compare their objective function values.
4
Evaluate the profit function Z=4S+9PZ = 4S + 9P at each vertex point.
At (60,40)(60, 40): Z=4(60)+9(40)=240+360=$600Z = 4(60) + 9(40) = 240 + 360 = \$600. At (120,0)(120, 0): Z=4(120)=$480Z = 4(120) = \$480. At (20,53)(20, 53): Z=4(20)+9(53)=80+477=$557Z = 4(20) + 9(53) = 80 + 477 = \$557. At (20,0)(20, 0): Z=4(20)=$80Z = 4(20) = \$80.
Direct substitution reveals which corner point yields the global maximum value inside the bounded constraints.

Key Concept

Linear programming optimization under joint resource and integer bound constraints
Question 14Question

A high-tech manufacturing firm produces custom drone components: Micro-Sensors (SS) and Control Units (UU). Each Micro-Sensor requires 2 hours of precision calibration and 1 GB of firmware memory allocation. Each Control Unit requires 5 hours of precision calibration and 4 GB of firmware memory allocation. For an upcoming production batch, the facility has a maximum of 40 hours of calibration time available and a maximum total memory allocation limit of 30 GB. If the firm earns a profit of 100perMicroSensorand100 per Micro-Sensor and 350 per Control Unit, which of the following combinations of Micro-Sensors (SS) and Control Units (UU) maximizes total batch profit while satisfying all operational bounds?

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Answer: 2 Micro-Sensors and 7 Control Units

Answer

The combination of 2 Micro-Sensors and 7 Control Units maximizes total profit ($2,650) under the given calibration and memory constraints.
The pair with 2 Micro-Sensors and 7 Control Units fully satisfies both constraints (2(2)+5(7)=39leq402(2) + 5(7) = 39 \\leq 40 hours and 2+4(7)=30leq302 + 4(7) = 30 \\leq 30 GB) and yields the maximum profit of $2,650.

Step-by-Step Solution

1
Set up the linear programming model with objective function and system of inequalities.
Maximize P=100S+350UP = 100S + 350U subject to: 2S+5Uleq402S + 5U \\leq 40 (Calibration limit), S+4Uleq30S + 4U \\leq 30 (Memory limit), and non-negative integers S,Ugeq0S, U \\geq 0.
Establishing explicit bounds is necessary to evaluate boundary integer points.
2
Solve for the intersection of the boundary equations 2S+5U=402S + 5U = 40 and S+4U=30S + 4U = 30.
Substituting S=304US = 30 - 4U into the first equation yields 2(304U)+5U=40implies603U=40impliesU=20/3approx6.672(30 - 4U) + 5U = 40 \\implies 60 - 3U = 40 \\implies U = 20/3 \\approx 6.67 and S=3.33S = 3.33.
The continuous optimal point occurs at the intersection of the binding constraints.
3
Evaluate feasible integer coordinates near the intersection boundary.
For U=7U = 7: Sleqmin(frac40352,3028)=min(2.5,2)=2implies(2,7)S \\leq \\min(\\frac{40 - 35}{2}, 30 - 28) = \\min(2.5, 2) = 2 \\implies (2, 7). Profit = 100(2)+350(7)=100(2) + 350(7) = \\ 2,650 .For. For U = 6 :: S \\leq \\min(\\frac{40 - 30}{2}, 30 - 24) = \\min(5, 6) = 5 \\implies (5, 6).Profit=. Profit = 100(5) + 350(6) = \\2,6002,600.
Since production quantities must be integers, testing feasible boundary pairs determines the global optimum.

Key Concept

Optimization under Bounded Linear Constraints
Question 15Question

A regional emergency health network deploys two types of mobile medical units for event coverage: Rapid Response Units (RR) and Heavy Support Units (HH).

The deployment is governed by the following staffing and operational constraints:
- Each Rapid Response Unit (RR) requires 11 paramedic and 22 EMTs.
- Each Heavy Support Unit (HH) requires 33 paramedics and 11 EMT.
- On any given shift, a maximum of 2121 paramedics and 2222 EMTs are available.
- Operational policy mandates deploying at least 33 Rapid Response Units (R3R \ge 3) and at least 22 Heavy Support Units (H2H \ge 2).

Each Rapid Response Unit can treat 1010 patients per hour, and each Heavy Support Unit can treat 1515 patients per hour.

Match each operational metric on the left to its corresponding optimal value on the right that maximizes total patient treatment capacity per hour.

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Items

Optimal number of Rapid Response Units (RR)
Optimal number of Heavy Support Units (HH)
Maximum total patient treatment capacity per hour

Matches

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Answer

The optimal operational configuration pairs 9 Rapid Response Units with 4 Heavy Support Units, achieving a maximum treatment capacity of 150 patients per hour.
The optimal pairing matches 9 Rapid Response Units, 4 Heavy Support Units, and a total capacity of 150 patients per hour. The optimal point (9,4)(9, 4) fully utilizes both available resource pools (1(9)+3(4)=211(9) + 3(4) = 21 paramedics and 2(9)+1(4)=222(9) + 1(4) = 22 EMTs), maximizing total throughput 10R+15H10R + 15H.

Step-by-Step Solution

1
Set up the linear inequality constraints and objective function.
Paramedic resource constraint: R+3H21R + 3H \le 21; EMT resource constraint: 2R+H222R + H \le 22; Operational bounds: R3,H2R \ge 3, H \ge 2; Objective function to maximize: C=10R+15HC = 10R + 15H.
Translating the narrative conditions into algebraic forms defines the feasible region of operation.
2
Calculate the corner vertices of the bounded feasible polygon.
Solving R+3H=21R + 3H = 21 and 2R+H=222R + H = 22 simultaneously yields R=9R = 9 and H=4H = 4. Other extreme vertices defined by lower bounds are (3,6)(3, 6), (3,2)(3, 2), and (10,2)(10, 2).
Linear programming optimization principles guarantee that the maximum objective value occurs at one of the corner vertices of the feasible region.
3
Evaluate the capacity objective function C(R,H)=10R+15HC(R, H) = 10R + 15H at each vertex.
At (3,2):10(3)+15(2)=60(3, 2): 10(3) + 15(2) = 60; At (3,6):10(3)+15(6)=120(3, 6): 10(3) + 15(6) = 120; At (10,2):10(10)+15(2)=130(10, 2): 10(10) + 15(2) = 130; At (9,4):10(9)+15(4)=150(9, 4): 10(9) + 15(4) = 150.
Comparing capacities across all vertices establishes (9,4)(9, 4) as the global maximum.

Key Concept

Linear Programming Optimization under Bounded System Constraints
Question 16Question

A renewable energy company designs micro-grid installations containing two types of modular components: Battery Storage Units (BB) and Solar Inverter Modules (II). The installation configuration is subject to the following system constraints:

1. Budget Bound: Each Battery Storage Unit costs \4,000andeachInverterModulecosts$2,000.Thetotalinstallationbudgetcannotexceed$32,000,givingtheconstraint4,000 and each Inverter Module costs \$2,000. The total installation budget cannot exceed \$32,000, giving the constraint 2B + I \le 16 .2.PeakPowerRequirement:Tomeetgridstabilityrequirements,thecombinedsystemratingmustsatisfy. 2. **Peak Power Requirement**: To meet grid stability requirements, the combined system rating must satisfy 3B + 4I \ge 30 .3.PhysicalSpaceLimit:Duetofloorloadcapacity,amaximumof6BatteryStorageUnitscanbeinstalled(. 3. **Physical Space Limit**: Due to floor load capacity, a maximum of 6 Battery Storage Units can be installed ( B \le 6$).

The total daily credit rating generated by the installation is given by the objective function E=12B+5IE = 12B + 5I.

Which combination of Battery Storage Units (BB) and Solar Inverter Modules (II) satisfies all system constraints while maximizing the total daily credit rating EE?

Show answer & explanation

Answer: 6 Battery Storage Units and 4 Inverter Modules

Answer

6 Battery Storage Units and 4 Inverter Modules
The combination of 6 Battery Storage Units and 4 Inverter Modules strictly satisfies all three constraints: the space constraint (666 \le 6), the budget constraint (2(6)+4=16162(6) + 4 = 16 \le 16), and the peak power requirement (3(6)+4(4)=34303(6) + 4(4) = 34 \ge 30). Substituting these values into the credit rating function yields E=12(6)+5(4)=92E = 12(6) + 5(4) = 92, which is higher than any other feasible combination.

Step-by-Step Solution

1
Identify the feasible values for BB based on bounded constraints.
The physical space constraint limits BB to non-negative integers where B6B \le 6.
Establishing individual bounds reduces the search space for integer solutions.
2
Evaluate upper bounds for II using the budget constraint I162BI \le 16 - 2B for candidate values of BB.
For B=6B = 6, I1612=4I \le 16 - 12 = 4. For B=5B = 5, I1610=6I \le 16 - 10 = 6. For B=4B = 4, I168=8I \le 16 - 8 = 8.
Determines maximum allowable Inverter Modules for each valid count of Battery Storage Units.
3
Check peak power constraint 3B+4I303B + 4I \ge 30 and compute objective function E=12B+5IE = 12B + 5I for feasible candidate pairs.
Pair (6,4)(6, 4): 3(6)+4(4)=34303(6) + 4(4) = 34 \ge 30, E=12(6)+5(4)=92E = 12(6) + 5(4) = 92.
Pair (5,6)(5, 6): 3(5)+4(6)=39303(5) + 4(6) = 39 \ge 30, E=12(5)+5(6)=90E = 12(5) + 5(6) = 90.
Pair (4,8)(4, 8): 3(4)+4(8)=44303(4) + 4(8) = 44 \ge 30, E=12(4)+5(8)=88E = 12(4) + 5(8) = 88.
Testing valid candidates against all joint constraints identifies the configuration yielding maximum credits.
4
Compare total daily credit ratings across all valid choices.
The maximum credit rating is 92, achieved at B=6B = 6 and I=4I = 4.
Confirms the optimal bounded solution.

Key Concept

Optimization under Bounded Linear Constraints
Estimated Time:2m 0s