CBSE Class 12 · 2006 · Linear ProgrammingLong Answer

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A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5,760 to invest and has space for at most 20 items. A fan and a sewing machine cost Rs. 360 and Rs. 240 respectively. He can sell a fan at a profit of Rs. 22 and a sewing machine at a profit of Rs. 18. Assuming that he can sell whatever he buys, how should he invest his money in order to maximize his profit? Translate the problem into an LPP and solve it graphically.

Answer: Max profit = Rs 392 at (8, 12): 8 fans and 12 sewing machines

Solution

Let the dealer buy

x

fans and

y

sewing machines.
Profit:

Z = 22 x + 18 y

.
Space:

x + y \leq 20

Investment:

360 x + 240 y \leq 5760 ⟹ 3 x + 2 y \leq 48

x \geq 0, y \geq 0

Corner points:

(0, 0)

(16, 0)

(8, 12)

(intersection of

x + y = 20

and

3 x + 2 y = 48

) and

(0, 20)

.
Evaluate

Z = 22 x + 18 y

(0, 0)

Z = 0

(16, 0)

Z = 352

(8, 12)

Z = 176 + 216 = 392

← maximum

(0, 20)

Z = 360

∴ Z_{m a x} = Rs 392 at (8, 12)

The dealer should buy 8 fans and 12 sewing machines.

CBSE Class 12 Board · 2006 · Subjective

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CBSE Class 12 · 2006 · Linear ProgrammingLong Answer

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