A linear programming problem is as follows: Maximize/minimize objective function $Z = 2 x - y + 5$ .
Subject to constraints
$3 x + 4 y \leq 60, x + 3 y \leq 30, x \geq 0, y \geq 0$
If the comer points of the feasible region are A(0, 10), B(12, 6), C(20, 0) and 0(0, 0), then which of the following is true?

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Difference of maximum and minimum values of Z is 35

At two comer points value of Z are equal.

Minimum value of Z is -5

Maximum value of Z is 40

answer is B.

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Detailed Solution

$Z = 2 x - y + 5$

Values of Z at comer points A, B, C and 0 are

at $A (0, 10), Z = 2 \times 0 - 10 + 5 = - 5 \to$ Minimum

at $B (12, 6), Z = 2 \times 12 - 6 + 5 = 23$

at $C (20, 0), Z = 2 \times 20 - 0 + 5 = 45 \to$ Maximum

at $O (0, 0), Z = 2 \times 0 - 0 + 5 = 5$ .

$∴$ Option (b) is the correct answer.

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