MATH SOLVE

2 months ago

Q:
# Suppose that the cost of producing x units is c(x)=50x dollars and that the revenue earned from selling x units is R(x)=-5x^2+100x dollars. Determine P(x), the profit in dollars from selling x units. Determine the number of units required to break even. If there are multiple break even points, list all of them. Determine the number of units required to maximize the profit. Also, determine the maximum profit. Justify why there is a maximum

Accepted Solution

A:

a] The profit function will be given as follows:

Profit=Revenue-Cost

P(x)=R(x)-C(x)

P(x)=-5x^2+100x-50x

P(x)=-5x^2+50x

b] The number of units at breakeven even point:

At breakeven point:

P(x)=0

hence:

R(x)=C(x)

thus

-5x^2+100x=50x

solving for x we get:

-5x^2=50x-100x

-5x^2=-50x

hence

x=10 units

c] The number of units that will maximize the profit will be found as follows:

P(x)=-5x^2+50x

b/-2a=50/(-2*-5)=5 units

D] the maximum profit will be:

x=5

P(x)=-5x^2+50x

P(5)=-5(5)^2+50*5

P(5)=125

Profit=Revenue-Cost

P(x)=R(x)-C(x)

P(x)=-5x^2+100x-50x

P(x)=-5x^2+50x

b] The number of units at breakeven even point:

At breakeven point:

P(x)=0

hence:

R(x)=C(x)

thus

-5x^2+100x=50x

solving for x we get:

-5x^2=50x-100x

-5x^2=-50x

hence

x=10 units

c] The number of units that will maximize the profit will be found as follows:

P(x)=-5x^2+50x

b/-2a=50/(-2*-5)=5 units

D] the maximum profit will be:

x=5

P(x)=-5x^2+50x

P(5)=-5(5)^2+50*5

P(5)=125