The weekly revenue for a product is given by Rx3762x0036x2 a

The weekly revenue for a product is given by R(x)=376.2x-0.036x², and the weekly cost is C(x)=9000+188.1x-0.072x²+0.00003x³, where x is the number of units produced and sold.

A. How many units will give the maximum profit?

B. What is the maximum possible profit?

Solution

P(x)=R(x)-C(x)=376.2x-0.036x² -(9000+188.1x-0.072x²+0.00003x³)

To maximize the profit lets find x for dP/dx=0

376.2-.072x-188.1+.144x-.00009x² =0

9x² -7200x-18810000=0

x² -800x-2090000=0

x=1900,-1100

d²p/dx² =.072-.00018x=T

If T>0 minima T<0 maxima

at x=1900 T=-.27

at x=-1100 T=.27

A.1900 units

B.376.2*1900-0.036*1900² -(9000+188.1*1900-0.072*1900²+0.00003*1900³) = 272580