Answer:
a. 1620-x^2
b. x=810
c. Maximum value revenue=$656,100
Step-by-step explanation:
(a) Total revenue from sale of x thousand candy bars
P(x)=162 - x/10
Price of a candy bar=p(x)/100 in dollars
1000 candy bars will be sold for
=1000×p(x)/100
=10*p(x)
x thousand candy bars will be
Revenue=price × quantity
=10p(x)*x
=10(162-x/10) * x
=10( 1620-x/10) * x
=1620-x * x
=1620x-x^2
R(x)=1620x-x^2
(b) Value of x that leads to maximum revenue
R(x)=1620x-x^2
R'(x)=1620-2x
If R'(x)=0
Then,
1620-2x=0
1620=2x
Divide both sides by 2
810=x
x=810
(C) find the maximum revenue
R(x)=1620x-x^2
R(810)=1620x-x^2
=1620(810)-810^2
=1,312,200-656,100
=$656,100
Answer:
a) [tex]R(x) = 162,000x-100x^2[/tex]
b) x = 810
c) 656,100$
Step-by-step explanation:
[tex]p(x) = 162- \frac{x}{10}[/tex]
a) find expression for total revenue from sale ox x thousand candy bars
R(x) = (x)[p(x)]
R(x) = x(162 - x/10)
therefore, [tex]R(x) = 162,000x-100x^2[/tex]
b) find value of x that leads to maximum revenue
take derivative of R(x), which is: [tex]R'(x) = 162,000-200x[/tex]
set that equation equal to 0 and solve for critical number(s): [tex]162,000-200x=0[/tex]
therfore, critical number is: x = 810
**this critical number is the x-value that leads to max. revenue
c) find the maximum revenue
take original R(x) formula, and insert the critical number you just solved for and insert it into there to solve for part c.
[tex]R(x) = 162,000x-100x^2[/tex]
[tex]R(810)=162,000(810)-100(810)^2[/tex]
therefore, your maximum revenue in dollars is: 656,100$
