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

The Correct Option is D

To solve this linear programming problem, we need to optimize \( Z = 3x + 9y \) subject to the following constraints:  

• \( x + 3y \leq 60 \)
• \( x + y \geq 10 \)
• \( x \leq y \)
• \( x \geq 0 \)
• \( y \geq 0 \)

The feasible region is determined by the intersection of these constraints.

Constraint	Boundary Line Equation
\( x + 3y \leq 60 \)	\( x + 3y = 60 \)
\( x + y \geq 10 \)	\( x + y = 10 \)
\( x \leq y \)	\( x = y \)
\( x \geq 0 \)	\( x = 0 \)
\( y \geq 0 \)	\( y = 0 \)

We find the feasible region by plotting these lines and identifying the intersection points:

• Solving \( x + 3y = 60 \) and \( x + y = 10 \): \( x = -15, y = 25 \) (out of feasible region)
• Solving \( x + 3y = 60 \) and \( x = y \): \( x = 15, y = 15 \)
• Solving \( x + y = 10 \) and \( x = y \): \( x = 5, y = 5 \) (not meeting all constraints)
• Solving \( x = 0 \) and \( x + 3y = 60 \): \( y = 20 \) (point \( (0, 20) \) )
• Solving \( x = 0 \) and \( x + y = 10 \): \( y = 10 \) (point \( (0, 10) \) )

Thus, the feasible region is bounded by points \( (0, 10) \), \( (0, 20) \), and \( (15, 15) \).

Calculating \( Z = 3x + 9y \) at vertices:

• At \( (0, 20) \): \( Z = 3(0) + 9(20) = 180 \)
• At \( (15, 15) \): \( Z = 3(15) + 9(15) = 180 \)
• At \( (0, 10) \): \( Z = 3(0) + 9(10) = 90 \)

The maximum value of \( Z \) is 180. Since \( Z = 180 \) at both \( (0, 20) \) and \( (15, 15) \), all points on the line segment joining these two will also yield \( Z = 180 \). Thus, the maximum value of \( Z \) occurs at all points on the segment between \( (0, 20) \) and \( (15, 15) \).

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.