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