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Home » A toy company manufactures two types of dolls, A and B. Each doll of type B take twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. the supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profit of Rs.3 and per dolls respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.
A toy company manufactures two types of dolls, A and B. Each doll of type B take twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. the supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profit of Rs.3 and \pm5 per dolls respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.
CBSE | Class 12 | Exercise 33B | Linear Programming | Maths | RS Aggarwal
A toy company manufactures two types of dolls, A and B. Each doll of type B take twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. the supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profit of Rs.3 and \pm5 per dolls respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.

Let x and y be number of doll A manufactured and doll B manufactured.
\therefore According to the question,
x+y \leq 1500, x+2 y \leq 2000, y \leq 600, x \geq 0, y \geq 0
Maximize Z=3 x+5 y
The feasible region determined by x+y \leq 1500, x+2 y \leq 2000, y \leq 600, x \geq 0, y \geq 0 is given by

The corner points of feasible region are A(0,0), B(0,600), C(800,600), D(1000,500), E(1500,0). The value of Z at corner points are

    \[\begin{tabular}{|l|l|l|} \hline Corner Point & $Z=3 x+5 y$ & \\ \hline $\mathrm{A}(0,0)$ & 0 & \\ \hline $\mathrm{B}(0,600)$ & 3000 & \\ \hline $\mathrm{C}(800,600)$ & 5400 & \\ \hline $\mathrm{D}(1000,500)$ & 5500 & Maximum \\ \hline $\mathrm{E}(1500,0)$ & 4500 & \\ \hline \end{tabular}\]

The maximum value of Z is 5500 at point (1000,500).
Hence, the manufacturer should produce 1000 types of doll A and 500 types of doll B to make maximum profit of Rs. 5500 .

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