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Revenue Maximization

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Demand for a certain product is represented by the following equation:
p=(80-q)/4; 0<=q<=80
Where q is the # of units and p is the price per unit. At what quantity sold will the revenue be maximized?

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The revenue function is R(q) = p * q = [(80 - q)/4] * q = (80q / 4) - (q^2 /4) = 20q - (q^2 / 4)

Now, we differentiate ...

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