Question:medium

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

Updated On: Jan 16, 2026
  • Maximum value of Z occurs at the point (15, 15) only.
  • Maximum value of Z occurs at the point (0, 20) only.
  • Maximum value of Z occurs exactly at two points (15, 15) and (0, 20).
  • Maximum value of Z occurs at all the points on the line segment joining (15, 15) and(0, 20).
Show Solution

The Correct Option is D

Solution and Explanation

This linear programming problem requires the optimization of \( Z = 3x + 9y \) 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 defined by the intersection of these inequalities.

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

The feasible region is determined by plotting these boundary lines and finding their intersection points:

  • Intersection of \( x + 3y = 60 \) and \( x + y = 10 \): \( x = -15, y = 25 \) (outside the feasible region)
  • Intersection of \( x + 3y = 60 \) and \( x = y \): \( x = 15, y = 15 \)
  • Intersection of \( x + y = 10 \) and \( x = y \): \( x = 5, y = 5 \) (does not satisfy all constraints)
  • Intersection of \( x = 0 \) and \( x + 3y = 60 \): \( y = 20 \), yielding point \( (0, 20) \)
  • Intersection of \( x = 0 \) and \( x + y = 10 \): \( y = 10 \), yielding point \( (0, 10) \)

The vertices defining the feasible region are \( (0, 10) \), \( (0, 20) \), and \( (15, 15) \).

The objective function \( Z = 3x + 9y \) is evaluated at each vertex:

  • 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 \( (15, 15) \), the maximum value of \( Z \) is achieved at all points along the line segment connecting \( (0, 20) \) and \( (15, 15) \).

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