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Let π = -20 + 120*Q – 3*Q2 + 10*ADV - .05*ADV2, where π = profit,...

Let π = -20 + 120*Q – 3*Q2 + 10*ADV - .05*ADV2, where π = profit, Q = units produced and sold, and ADV = units of advertising. What level of Q maximizes profits?

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Answer #1

Ans. Profit (P) = -20 + 120Q -3Q^2 + 10ADV - 0.05ADV^2

To maximize profit, differentiate profit with respect to Q and equate the result to zero,

dP/dQ = 120 - 6Q = 0

=> Q = 20 units

Thus, at Q = 20 units, profit is maximized.

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