Name: Linear Programming: Manufacturing Problem
Uploaded: 2019-07-19
Duration: 7 min 3 s
Description: In this video we are going to learn linear programming manufacturing problem comprehensively with the help of real life examples. Check it out!

Question

Minimize $z = - 3 x + 3 y$ subject to $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$ .

Accepted Answer

Given,

Minimise $Z = - 3 x + 3 y$ , subject to $x + 2 y \leq 8$ , $3 x + 2 y \leq 12$ , $x \geq 0, y \geq 0$

Points to plot graph for $x + 2 y = 8$ are taken as below.

$\begin{array}{l} x & 0 & 8 \\ y & 4 & 0 \end{array}$

Points to plot graph for $3 x + 2 y = 12$ are taken as below.

$\begin{array}{l} x & 0 & 4 \\ y & 6 & 0 \end{array}$

For point of intersection of the lines $x + 2 y = 18 . . . . (i)$ and $3 x + 12 y = 12 . . . . . . (i i)$ .

Solving the above equations, by subtracting $(i i) - 3 (i)$ , we get

$(3 x + 2 y) - 3 (x + 2 y) = 12 - 3 (8)$

$\Rightarrow 2 y - 6 y = 12 - 24$

$\Rightarrow - 4 y = - 12$

$\Rightarrow y = 3$

Substituting $y = 3$ in any one of the equation, we get $x = 2$ .

Therefore, intersection point of two lines is $(2, 3)$

The corner points for $Z = - 3 x + 3 y$ are $O (0, 0), A (0, 4), B (2, 3), C (4, 0)$ .

At $O (0, 0), Z = 0$

At $A (0, 4), Z = 12$

At $B (2, 3), Z = 3$

At $C (4, 0), Z = - 12$

Therefore, minimum value of $Z = - 12$ is obtained at $(4, 0)$ .

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