Probem54+Elise+David+AC

54. 

P(x) = R(x) - C(x) R(x)=xp(x) marginal cost=C'(x) marginal revenue = R'(x)

a.)P(x)=R(x) - C(x) P'(x)=R'(x)-C'(x)=0 R'(x)=C'(x) you find a maximum value by setting the derivative equal to zero, sowhen the marginal cost equals the marginal revenue, the profit is at a maximum

b.) C(x)=1600+500x-1.6x^2+0.004x^3 P(x)=1700-7x R(x)=x(1700-7x)=1700x-7x^2

To maximize P, R'(x)=C'(x) so: 1700-14x=500-3.2x+0.012x^2, x>0 0.012x^2+10.8x-1200=0 x=100, -1000 -1000 not in the domain Therefore the production level that will maximize profit is 100 units