Question

Difficulty: HardSystems of Linear Inequalities in Two Variables

A beverage company produces a fruit juice blend using apple juice and cranberry juice. Let xx represent the volume, in liters, of apple juice, and let yy represent the volume, in liters, of cranberry juice used in one batch of the blend. The production constraints for each batch are modeled by the system of inequalities below:

x+y100x + y \geq 100
y13xy \geq \frac{1}{3}x
y12x+75y \leq -\frac{1}{2}x + 75

What is the maximum possible volume, in liters, of cranberry juice that can be used in a single batch?

Answer: 50 liters

Answer

50
To find the maximum possible value of yy, we examine the bounds of the feasible region. Rewriting the inequalities in terms of xx gives x100yx \geq 100 - y and x1502yx \leq 150 - 2y. Since xx must satisfy both inequalities simultaneously, the lower bound must be less than or equal to the upper bound, meaning 100y1502y100 - y \leq 150 - 2y. Solving this inequality for yy gives y50y \leq 50. The point (50,50)(50, 50) also satisfies the remaining constraint y13xy \geq \frac{1}{3}x, confirming that 50 is the maximum possible value.

Step-by-Step Solution

1
Isolate xx in the first inequality: x+y100x + y \geq 100 becomes x100yx \geq 100 - y.
x100yx \geq 100 - y
This establishes a lower bound for the volume of apple juice in terms of the volume of cranberry juice.
2
Isolate xx in the third inequality: y12x+75y \leq -\frac{1}{2}x + 75 can be rewritten by subtracting 75 from both sides to get y7512xy - 75 \leq -\frac{1}{2}x, then multiplying by 2-2 and reversing the inequality sign to get x1502yx \leq 150 - 2y.
x1502yx \leq 150 - 2y
This establishes an upper bound for the volume of apple juice in terms of the volume of cranberry juice.
3
Combine the lower and upper bounds of xx: 100yx1502y100 - y \leq x \leq 150 - 2y. For a valid solution to exist, the lower bound must be less than or equal to the upper bound.
100y1502y100 - y \leq 150 - 2y
This allows us to solve directly for the constraint on the cranberry juice volume yy.
4
Solve the inequality 100y1502y100 - y \leq 150 - 2y by adding 2y2y to both sides and subtracting 100 from both sides.
y50y \leq 50
This determines that the maximum possible value for yy is 50.
5
Substitute y=50y = 50 back into the compound inequality to find xx: 10050x1502(50)    50x50100 - 50 \leq x \leq 150 - 2(50) \implies 50 \leq x \leq 50, which means x=50x = 50. Check if the point (50,50)(50, 50) satisfies the remaining inequality y13xy \geq \frac{1}{3}x: 5013(50)16.6750 \geq \frac{1}{3}(50) \approx 16.67, which is true.
The point (50,50)(50, 50) is a valid solution to the system.
This confirms that 50 is a achievable value within the system of inequalities.

Key Concept

Finding the boundary limits and optimization points within a system of linear inequalities in two variables.
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