MHT CET · Maths · Linear Programming
A manufacturing company produces two items, A and B. Each toy should be processed by two machines, I and II. Machine I can be operated for maximum 10 hours 40 minutes. It takes 20 minutes for an item A and 15 minutes for B. Machine II can be operated for a total time at 8 hours 20 minutes. It takes 5 minutes for an item \(A\) and 8 minutes for \(B\). The profit per item of \(A\) is ₹ \(25\) and per item of \(B\) is ₹ 18. The formulation of an L.P.P. to maximize the profit (where \(x\) is number of items \(A\) and \(y\) is the number of item \(B\) ) is _____
- A Maximize \(z=25 x+18 y\) subject to \(20 x+15 y \leqslant 640\) \(5 x+8 y \geqslant 500\) \(x, y \geqslant 0\)
- B Maximize \(z =25 x+18 y\) subject to \(20 x+15 y \leqslant 640\) \(5 x+8 y \leqslant 500\) \(x, y \geqslant 0\)
- C Maximize \(z=25 x+18 y\) subject to \(20 x+5 y \leqslant 8\) \(5 x+8 y \leqslant 10\) \(x, y \geqslant 0\)
- D Maximize \(z =25 x+18 y\) subject to \(4 x+3 y \leqslant 128\) \(5 x+8 y \geqslant 500\) \(x, y \geqslant 0\)
Answer & Solution
Correct Answer
(B) Maximize \(z =25 x+18 y\) subject to \(20 x+15 y \leqslant 640\) \(5 x+8 y \leqslant 500\) \(x, y \geqslant 0\)
Step-by-step Solution
Detailed explanation
Maximize \(z = 25x + 18y\) Subject to:
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