A printing company produces two types of customized marketing materials: Standard brochure… — GMAT

A printing company produces two types of customized marketing materials: Standard brochures ( $S$ ) and Premium brochures ( $P$ ). Each Standard brochure requires $2$ minutes of printing time and $1$ minute of binding time. Each Premium brochure requires $3$ minutes of printing time and $3$ minutes of binding time. The production facility has a maximum daily capacity of $240$ minutes for printing and $180$ minutes for binding. Due to a recurring client agreement, the facility must produce at least $20$ Standard brochures per day. The profit is $\$4$ per Standard brochure and $\$9$ 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.

Optimal number of Standard brochures ( $S$ ) $60$
Optimal number of Premium brochures ( $P$ ) $40$
Maximum total daily profit ( $Z$ ) $\$600$

Cevap

The optimal production values are

60

Standard brochures and

40

Premium brochures, resulting in a maximum daily profit of

\$600

Solving the system of simultaneous constraints yields the optimal vertex at

S = 60

Standard brochures and

P = 40

Premium brochures. At this point, both printing (

2(60) + 3(40) = 240

) and binding (

60 + 3(40) = 180

) capacities are fully utilized, achieving the maximum daily profit of

4(60) + 9(40) = \$600

Adım Adım Çözüm

Formulate the linear programming model with constraints and objective function.

Constraints:

2S + 3P \le 240

(printing time),

S + 3P \le 180

(binding time),

S \ge 20

(contractual minimum), and

S, P \ge 0

. Objective: Maximize

Z = 4S + 9P

Clear mathematical formulation is required to define the feasible region bounded by resource limits.

Calculate the intersection points (vertices) of the boundary lines of the feasible region.

Subtracting

S + 3P = 180

from

2S + 3P = 240

yields

S = 60

. Substituting

S = 60

into

S + 3P = 180

gives

3P = 120 \implies P = 40

. Corner point 1:

(60, 40)

The fundamental theorem of linear programming states that maximum profit occurs at a vertex of the feasible region.

Find remaining boundary vertices satisfying all constraints.

Corner point 2:

(120, 0)

from printing limit when

P=0

(satisfies binding limit

120 \le 180

). Corner point 3:

(20, 53)

when

S=20

(integer constraint on

P

where

20 + 3P \le 180 \implies P \le 53.33

). Corner point 4:

(20, 0)

All feasible vertices must be identified to compare their objective function values.

Evaluate the profit function

Z = 4S + 9P

at each vertex point.

(60, 40)

Z = 4(60) + 9(40) = 240 + 360 = \$600

. At

(120, 0)

Z = 4(120) = \$480

. At

(20, 53)

Z = 4(20) + 9(53) = 80 + 477 = \$557

. At

(20, 0)

Z = 4(20) = \$80

Direct substitution reveals which corner point yields the global maximum value inside the bounded constraints.

Anahtar Kavram

Linear programming optimization under joint resource and integer bound constraints

Soru

Cevap

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