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Please help me solve C and D Please!!!

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Oct 19th, 2017

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) = $6,666.67

c)  R(x) = -x^2/3 +100x 

differentiate w.r.t  x
R'(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

Mar 30th, 2015

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