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Graphical Method of Solving Linear Programming Problems

A factory man...

Question

A factory manufactures two types of screws A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 min on the automatic and 6 min on hand operated machines to manufacture a package of screws A, while it takes 6 min on automatic and 3 min on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 h on any day. The manufacturer can sell a package of screws A at a profit on Rs. 7 and screws B at a profit of Rs.10. Assuming that he can sell all the screws he manufactured, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

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Solution

Let the manufacturer produces x package of screws A and y package of screws B, We construct the following table:

$\begin{matrix} T y p e o f s c r e w & N u m b e r o f & T i m e o n a u t o m a t i c & T i m e o n h a n d & P r o f i t p a c k a g e s & m a c h i n e (i n m i n) & m a c h i n e (i n m i n) & (i n R s .) A & x & 4 x & 6 x & 7 x B & y & 6 y & 3 y & 10 y T o t a l & x + y & 4 x + 6 y & 6 x + 3 y & 7 x + 10 y A v a i l a b i l i t y & 4 \times 60 = 240 & 4 \times 60 = 240 \end{matrix}$

The profits on a package of screws A is Rs. 7 and on the package of screws B is Rs. 10.

Our problem is to maximize $Z = 7 x + 10 y . . . . . . . . . . (i)$

Subject to constraints $4 x + 6 y \leq 240 \Leftrightarrow 2 x + 3 y \leq 120 . . . . . . . . . (i i)$

$6 x + 3 y \leq 240 \Leftrightarrow 2 x + y \leq 80 . . . . . . . . . (i i i)$

and $x \geq 0, y \geq 0 . . . . . . . . . . (i v)$

Firstly, draw the graph of the line 2x + 3y = 120

$\begin{matrix} x & 0 & 60 y & 40 & 0 \end{matrix}$

Putting (0, 0) in the inequality

$2 x + 3 y \leq 120$ , we have $2 \times 0 + 3 \times 0 \leq 120$

$\Rightarrow 0 \leq 120$ (which is true)

So, the half plane is towards the origin. Secondly, draw the graph of the line $2 x + y = 80$

$\begin{matrix} x & 40 & 0 y & 0 & 80 \end{matrix}$

Putting (0, 0) in the inequality $2 x + y \leq 80$ , we have $2 \times 0 + 0 \leq 80 \Rightarrow 0 \leq 80$ (which is true)

So, the half plane is towards the origin. Since, x, y $\geq$ 0

So, the feasible region lies in the first quadrant.

On solving equations $2 x + 3 y = 120 a n d 2 x + y = 80,$ we get B(30, 20).

The corner points of the feasible region are O(0, 0), A(40, 0), B(30, 20) and C(0, 40). the values of Z at these points are as follows:

$∴$ Feasible region is OABCO.

$\begin{matrix} C o r n e r p o i n t & Z = 7 x + 10 y O (0, 0) & 0 A (40, 0) & 280 B (30, 20) & 410 \to M a x i m u m C (0, 40) & 400 \end{matrix}$

The maximum value of Z is Rs. 410 at B(30, 20)

Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs. 410.

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