A researcher is preparing a growth medium for a bacterial culture by mixing Nutrient A and… — SAT

A researcher is preparing a growth medium for a bacterial culture by mixing Nutrient A and Nutrient B. Let $x$ represent the number of grams of Nutrient A and $y$ represent the number of grams of Nutrient B in the mixture. The mixture must satisfy the following conditions:

* The total mass of the nutrients in the mixture is at most $15$ grams.
* The mass of Nutrient B is at most twice the difference of the mass of Nutrient A and $2$ grams.
* The mass of Nutrient B is at least half the mass of Nutrient A.

What is the maximum possible integer number of grams of Nutrient B that can be used in the mixture?

A
5
B
6
8Answer
D
9

Answer

The maximum possible integer number of grams of Nutrient B that can be used is 8.

The system of inequalities bounding the feasible region is

x + y \leq 15

y \leq 2x - 4

, and

y \geq \frac{1}{2}x

. The maximum value of

y

(representing Nutrient B) occurs at the intersection of the boundary lines

x + y = 15

and

y = 2x - 4

. Solving this system gives

x = 19/3

and

y = 26/3 \approx 8.67

. Since the researcher needs an integer number of grams, the largest integer value of

y

that falls within the region is

8

. Setting

y = 8

yields the valid range

6 \leq x \leq 7

, confirming that

y = 8

is a possible value.

Step-by-Step Solution

Translate the verbal constraints into a system of linear inequalities.

The system of inequalities is:
1)

x + y \leq 15

y \leq 2(x - 2) \implies y \leq 2x - 4

y \geq \frac{1}{2}x

Establishing the mathematical model is necessary to define the feasible region in the coordinate plane.

Find the intersection points of the boundary lines to locate the vertices of the feasible region.

Vertex 1 (Intersection of lines 1 and 2):

x + (2x - 4) = 15 \implies 3x = 19 \implies x = 19/3

, so

y = 26/3

.

Vertex 2 (Intersection of lines 1 and 3):

x + \frac{1}{2}x = 15 \implies \frac{3}{2}x = 15 \implies x = 10

, so

y = 5

.

Vertex 3 (Intersection of lines 2 and 3):

2x - 4 = \frac{1}{2}x \implies \frac{3}{2}x = 4 \implies x = 8/3

, so

y = 4/3

The maximum value of a coordinate in a bounded linear inequality system occurs at one of the vertices of the feasible region.

Identify the vertex that maximizes the

y

-coordinate.

Comparing the

y

-coordinates of the vertices:
-

y = 4/3 \approx 1.33

(8/3, 4/3)

y = 5

(10, 5)

y = 26/3 \approx 8.67

(19/3, 26/3)

The maximum value of

y

in the feasible region is

26/3 \approx 8.67

Comparing the coordinates determines the upper limit of the variable

y

in the shaded region.

Find the maximum possible integer value for

y

within the feasible region.

The maximum value is

8.67

. The largest integer less than or equal to

8.67

8

. Verify that

y = 8

is achievable by substituting it into the inequalities:

x + 8 \leq 15 \implies x \leq 7

8 \leq 2x - 4 \implies 2x \geq 12 \implies x \geq 6

8 \geq \frac{1}{2}x \implies x \leq 16

This yields the interval

6 \leq x \leq 7

, which contains valid solutions (such as the integer points

(6, 8)

and

(7, 8)

Ensures that the integer value chosen actually lies within the set of valid coordinates for the system.

Key Concept

Maximizing a variable under a system of linear constraints

Estimated Time:3m 0s

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