p = (-1/3)x + 100 , 0<= x <= 300 Revenue = price x quantity sold R = px = [(-1 3)x +100]x a) R(x) = -x^2/3 +100x b) If 100 units were sold: R(x) = -100^2 / 3 + 100(100) = $6,666.67c) R(x) = -x^2/3 +100x

differentiate w.r.t xR'(x) = -2x/3 +100 = 0 2x/3 = 100 2x=300 x=150

A quantity of x=150 maximizes the revenue beccause [R''(x) = -2/3 < 0] d) R(150) = -x^2/3 +100x = -22500/3 + 100*150 = -7500+15000 = $7500

Maximum Revenue will be at x = 150 is $7500

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