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Suppose that price is given by P = 481ecQ ; where Q denotes the quantity sold and cis a constant. Furthermore suppose that revenue is maximised at Q = 187: Find the value of c

Question

Suppose that price is given by P = 481ecQ ; where Q denotes the quantity sold and cis a constant. Furthermore suppose that revenue is maximised at Q = 187: Find the value of c

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Solution

The revenue R is given by the product of the price P and the quantity Q, i.e., R = P*Q.

Given that P = 481ecQ, we can write R as R = 481ecQ * Q = 481ecQ^2.

We know that the revenue is maximized at Q = 187. At the maximum point, the derivative of the revenue function with respect to Q should be zero.

So, let's differentiate R with respect to Q and set it equal to zero.

dR/dQ = 0 = 481ecQ^2 * 2Q

Solving for c, we get:

0 = 962ecQ^3

Since Q is not zero, we can divide both sides by 962eQ^3 to get:

c = 0

So, the value of c that maximizes revenue is 0.

This problem has been solved

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