A manufacturer make two types of toys A and B- Three machines are needed for this purpose and the time in minutes required for each toy on the machines is given below-Types - - Machines - Whline- - I - II - IIINlhineA - 12 - 18 - 6 1B - 6 - 0 - 9 Wlline array Each machine is available for a maximum of 6 h per day- If the profit on each toy of type A is Rs- 7-50 and that the each toy of type B is Rs- 5 - show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit-

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Optimization Problems

A manufacture...

Question

A manufacturer make two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below.
$\begin{matrix} T y p e s & M a c h i n e s I & I I & I I I A & 12 & 18 & 6 B & 6 & 0 & 9 \end{matrix}$
Each machine is available for a maximum of 6 h per day. If the profit on each toy of type A is Rs.7.50 and that the each toy of type B is Rs. 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

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Solution

Let the manufacturer makes x toys of type A and y toys of type B. We construct the following table:
$\begin{matrix} T y p e o f T o y s & N u m b e r o f t o y s & T i m e o f m a c h i n e I (i n m i n) & T i m e o f m a c h i n e I I (i n m i n) & T i m e o f m a c h i n e I I I (i n m i n) & P r i f i t (i n R s) A & x & 12 x & 18 x & 6 x & 7.50 x B & y & 6 y & 0 y & 9 y & 5 y T o t a l & x + y & 12 + 6 y & 18 x + 0 y & 6 x + 9 y & 7.50 x + 5 y M a x i m u m r e q u i r e s & 6 \times 60 = 360 & 6 \times 60 = 360 & 6 \times 60 = 360 \end{matrix}$
Our problem is to maximize Z = 7.50 x + 5y . . .(i)
Subject to constraints 12x + 6y $\leq$ 360 $\Leftrightarrow$ 2x + y $\leq$ 60 . . . (ii)
$18 x \leq 360 \Leftrightarrow x \leq 20$ . . .(iii)
$6 x + 9 y \leq 360 \Leftrightarrow 2 x + 3 y \leq 120$ . . . (iv)
$x \geq 0, y \geq 0$ . . . (v)
Firstly, draw the graph of the line 2x + y = 60
$\begin{matrix} x & 0 & 30 y & 60 & 0 \end{matrix}$
Putting (0, 0) in the inequality 2x + y $\leq$ 60, we have
$2 \times 0 + 0 \leq 60 \Rightarrow 0 \leq 60$ (which is true)
So, the half plane is towards the origin.
Secondly, 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 2x + 3y $\leq$ 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.
Thirdly, draw the graph of the line x = 20
Putting (0, 0) in the inequality $x \leq 20$ , we have $0 \leq 20$ (which is true)
So, the half plane is towards the origin.
Since, $x, y \leq 0$
So, the feasible region lies in the first quadrant.
On solving equations 2x + y = 60 and 2x + 3y = 120, we get C(15, 30)
Similarly, solving the equations x = 20 and 2x + y = 60, we get B(20, 20)
$∴$ Feasible region is OABCDO.
The corner points of the feasible region are A(20, 0), B(20, 20), C(15, 30) and D(0, 40).
The values of Z at these points are as follows:
$\begin{matrix} C o r n e r p o i n t & Z = 7.50 x + 5 y 0 (0, 0) & 0 A (20, 0) & 150 B (20, 20) & 250 C (15, 30) & 712.5 \to M a x i m u m D (0, 40) & 200 \end{matrix}$
Thus, the maximum value of Z is 712.5 C(15, 30).
Thus, manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.

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Q. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

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	I	II	III
A	12	18	6
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Q. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

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	I	II	III
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Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹7.50 and that on each toy of type B is ₹5, show that 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.

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