Solve the following Linear Programming problems graphically:
1. Maximize $Z = 3 x + 4 y$ Subject to the constraints : $x + y \leq 4, x \geq 0, y \geq 0$
2. Minimize $Z = - 3 x + 4 y$ subject to $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$

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Solution

1. Maximize $Z = 3 x + 4 y$

Corner points

Value of $Z = 3 x + 4 y$

(0,4)

16

(4,0)

9

(0,0)

0

Hence, Z is maximum at $(0, 4)$ and max value is $16$

2. Minimize $Z = - 3 x + 4 y$

Corner points

Value of $Z = - 3 x + 4 y$

(0,4)

16

(2,3)

6

(4,0)

-12

Hence, Z is minimum at $(4, 0)$ and min value is $- 12$

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Similar questions

z = 3 x + 4 y

x + y \leq 4, x \geq 0, y \geq 0

Z = - 3 x + 4 y

x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0

Minimize Z = -3x + 4y, subject to constraints are $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0 a n d y \geq 0$ .

Z = 200 x + 500 y . . . (1)

x + 2 y \geq 10 . . . (2)

3 x + 4 y \leq 24 . . . (3)

x \geq 0, y \geq 0 . . . (4)

Z = 3 x + 4 y

x + y \leq 4, x \geq 0, y \geq 0

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