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in Linear Programming by (25.7k points)
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A small firm manufactures items A and B. The total number of items that it can manufacture in a day is at most 24. Item A takes one hour to make while item B take only half an hour. The maximum time available per day is 16 hours. If the profit on one unit item A be ₹300 and that on one unit of item B be ₹160, how many of each type of item should be produced to maximize the profit? Solve the problem graphically.

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Best answer

Let the firm manufacture x number of A and y number of B products.

∴According to the question,

X + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0, y ≥ 0

Maximize Z = 300x + 160y 

The feasible region determined X + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0, y ≥ 0 is given by

The corner points of feasible region are A(0,0) , B(0,24) , C(8,16) , D(16,0).

The value of Z at corner point is

Corner Point Z = 300x + 160y
A(0, 0) 0
B(0, 24) 3840
C(8, 16) 4960 Maximum
D(16, 0) 4800

The maximum value of Z is 4960 and occurs at point (8,16).

The firm should produce 8 A products and 16 B products to earn maximum profit of Rs.4960.

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