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Home » A brick manufacture has two depots, and , with stocks of 30000 and 20000 bricks respectively. He receives order from three building , for 15000,20000 and 15000 bricks respectively. The costs of transporting 1000 bricks to the building from the depots are given below. How should the manufacture fulfill the orders so as to keep the cost of transportation minimum?
A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000 bricks respectively. He receives order from three building A, B, C, for 15000,20000 and 15000 bricks respectively. The costs of transporting 1000 bricks to the building from the depots are given below. How should the manufacture fulfill the orders so as to keep the cost of transportation minimum?
CBSE | Class 12 | Exercise 33B | Linear Programming | Maths | RS Aggarwal
A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000 bricks respectively. He receives order from three building A, B, C, for 15000,20000 and 15000 bricks respectively. The costs of transporting 1000 bricks to the building from the depots are given below. How should the manufacture fulfill the orders so as to keep the cost of transportation minimum?

Let x bricks be transported from P to A and y bricks be transported from P to B.
Therefore, 30000-(x+y) will be transported to C.
Also, (15000-x) bricks, ( 20000-y) bricks and (15000-(30000-(x+y))) bricks will be transported to A, B, C from Q.
\therefore According to the question,
x \geq 0, y \geq 0, x+y \leq 30000, x \leq 15000, y \leq 20000, x+y \geq 15000
Minimize Z=0.04 x+0.02(15000-x)+0.02 y+0.06(20000-y)+0.03(30000-(x+y))+0.04((x+y)- 15000)
Z=0.03 x-0.03 y+1800
The feasible region represented by x \geq 0, y \geq 0, x+y \leq 30000, x \leq 15000, y \leq 20000, x+y \geq 15000 is given by

The corner points of feasible region are A(0,15000), B(0,20000), C(10000,20000), D(15000,15000), E(15000,0)

    \[\begin{tabular}{|l|l|l|} \hline Corner Point & $\mathrm{Z}=0.03 \mathrm{x}-0.03 \mathrm{y}+1800$ & \\ \hline $\mathrm{A}(0,15000)$ & 1350 & \\ \hline $\mathrm{B}(0,20000)$ & 1200 & \\ \hline $\mathrm{C}(10000,20000)$ & 1500 & \\ \hline $\mathrm{D}(15000,15000)$ & 1800 & \\ \hline $\mathrm{E}(15000,0)$ & 2250 & \\ \hline \end{tabular}\]

The minimum value of Z is 1200 at point (0,20000).
Hence, 0,20000,10000 bricks should be transported from P to A, B, C and 15000,0,5000 bricks shoull transported from Q to A, B, C.

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