Question

Consider Max. z = - 2x - 3y subject to $\frac{x}{2} + \frac{y}{3} \leq 1 , \frac{x}{3} \frac{y}+{2} \leq 1 , x ,y \geq 0$ The max value of z is :

• A. 0
• B. 4
• C. 9
• D. 6

Answer 1

Answer: (A)

0

Answer 2

Given problem is max z = - 2x - 3y Subject to $\frac{x}{2} + \frac{y}{3} \leq 1 , \frac{x}{3}+ \frac{y}{2} \leq 1 , x ,y \geq 0$ First convert these inequations into equations we get 3x + 2y = 6 ...(i) 2x + 3y = 6 ...(ii) on solving these two equation, we get point of intersection is $\bigg( \frac{6}{5}, \frac{6}{5} \bigg)$ Now, we draw the graph of these lines. Shaded portion shows the feasible region. Now, the corner points are $(0, 2), (2,0), \bigg( \frac{6}{5}, \frac{6}{5} \bigg) , (0, 0).$ At (0, 2), value of z = - 2(0) - 3(2) = - 6 At (2, 0), value of z = - 2(2) - 3(0) = - 4 At $\bigg( \frac{6}{5}, \frac{6}{5} \bigg) , (0, 0).$ Value of $z = - 2 \left(\frac{6}{5} \right) - 3 \left(\frac{6}{5} \right)$ $= \frac{-30}{5} = - 6$ At (0, 0), value of z = - 2(0) - 3(0) = 0 At (0, 0), value of z = - 2(0) - 3(0) = 0 $\therefore$ The max value of z is 0.