Given:
The demand function is
[tex]p=-\frac{1}{20}x+920[/tex]The cost function is
[tex]C(x)=80x+6000[/tex]a)
Required:
We need to find the revenue function.
Explanation:
Multiply the demand function by x to find the revenue function.
[tex]R(x)=xp[/tex][tex]Substitute\text{ }p=\frac{-1}{20}x+920\text{ in the equation.}[/tex][tex]R(x)=x(\frac{-1}{20}x+920)[/tex][tex]R(x)=\frac{-1}{20}x^2+920x[/tex]Answer:
[tex]R(x)=-\frac{1}{20}x^2+920x[/tex]b)
Required:
We need to find the profit.
Explanation:
Recall that Profit=Revenue -Cost.
[tex]P(x)=R(x)-C(x)[/tex][tex]Substitute\text{ }R(x)=\frac{-1}{20}x^2+920x\text{ and }C(x)=80x+6000\text{ in the equation.}[/tex][tex]P(x)=\frac{-1}{20}x^2+920x\text{ -\lparen}80x+6000)[/tex][tex]P(x)=\frac{-1}{20}x^2+920x-80x-6000[/tex][tex]P(x)=\frac{-1}{20}x^2+840x-6000[/tex]Answer:
[tex]P(x)=\frac{-1}{20}x^2+840x-6000[/tex]c)
Required:
We need to find the maximum profit.
Explanation:
Consider the profit function.
[tex]P(x)=\frac{-1}{20}x^2+840x-6000[/tex]Differentiate this function with respect to x.
[tex]P^{\prime}(x)=\frac{-1}{20}(2x)+840(1)-6000(0)[/tex][tex]P^{\prime}(x)=\frac{-1}{10}x+840[/tex]Set P'(x) =0 to find the value of x that maximizes profit.
[tex]\frac{-1}{10}x+840=0[/tex]Subtract 840 from both sides.
[tex]\frac{-1}{10}x+840-840=0-840[/tex][tex]\frac{-1}{10}x=-840[/tex]Multiply both sides by (-10).
[tex]\frac{-1}{10}x(-10)=-840(-10)[/tex][tex]x=8400[/tex]Substitute x =8400 in the profit function to find the maximum profit.
[tex]P(8400)=\frac{-1}{20}(8400)^2+840(8400)-6000[/tex][tex]P(8400)=3522000[/tex]Answer:
The value of x that maximizes profit is 8400.
The maximum profit is $3,522,000.
d)
Required:
We need to find the price charged to maximize the profit.
Explanation:
Substitute x =8400 in the demand function.
[tex]p=-\frac{1}{20}(8400)+920[/tex][tex]p=500[/tex]Answer:
The price charged to maximize the profit is $500.
