Question

Optimize $Z = 3x + 9y$ subject to the constraints: $x + 3y \leq 60, \quad x + y \geq 10, \quad x \leq y, \quad x \geq 0, \quad y \geq 0.$

• A. Maximum value of Z occurs at the point (15, 15) only.
• B. Maximum value of Z occurs at the point (0, 20) only.
• C. Maximum value of Z occurs exactly at two points (15, 15) and (0, 20).
• D. Maximum value of Z occurs at all the points on the line segment joining (15, 15) and(0, 20).

Answer 1

Answer: (D)

Maximum value of Z occurs at all the points on the line segment joining (15, 15) and(0, 20).

Answer 2

To find the maximum value of $Z = 3x + 9y$, we graph the constraints: $x + 3y \leq 60$ represents a half-plane below the line $x + 3y = 60$. $x + y \geq 10$ represents a half-plane above the line $x + y = 10$. $x \leq y$ represents the region below the line $x = y$. $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant.

The feasible region is a polygon bounded by these lines. Evaluating $Z$ at the corner points of this polygon: At $(15, 15)$,

$Z = 3(15) + 9(15) = 180.$

At $(0, 20)$,

$Z = 3(0) + 9(20) = 180.$

Since $Z$ is linear and the line segment joining $(15, 15)$ and $(0, 20)$ lies within the feasible region, the maximum value of $Z$ occurs at all points on this line segment.