A manufacturer offers price per unit based on, order quantity “x” as per following function: p = 20 – 0.001x . The cost of manufacture for x units is given by the function: C = 5x + 2000....

A manufacturer offers price per unit based on, order quantity “x” as per following function: p = 20 – 0.001x . The cost of manufacture for x units is given by the function: C = 5x + 2000. Determine the number units he should make and sell per week to maximize the profit.

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p(x)=20-.001x

C(x)=5x+2000

R(x)=revenue function= x.p(x)=x(20-.001x)

=20x-0.001x^2

P(x)=profit function

P(x)=R(x)-C(x)

P(x)=20x-0.001x^2-5x-2000

P(x)=15x-0.001x^2-2000

differntiate P(x) ,w.r.t. x

P'(x)=15-0.002x

for maximum profit P'(x)=0

15-0.002x

x=7500

P''(x)=-0.002 <0

Thus x=7500 will give max profit.

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