The CarryItAll​ mini-van, a popular vehicle among soccer​ moms, obeys the demand equation

Question

ppequals=negative StartFraction 1 Over 40 EndFraction x plus 19 comma 000−140x+19,000.
The cost of producing
xx
vans is given by the function Upper C left parenthesis x right parenthesis equals 15450 x plus 20 comma 000.C(x)=15450x+20,000.
​a) Express the revenue
Upper RR
as a function of
xx.

​b) Express the profit
Upper PP
as a function of
xx.

​c) Find the value of
xx
that maximizes profit. What is the maximum​ profit?
​d) What price should be charged in order to maximize​ profit?

Answer 1

why all the words, when you have the symbols?

demand quantity at price x is
p(x) = 1/40 x + 19000
The cost is
C(X) = 15450x + 20000
(a)
So, the revenue (price * quantity) is
R(X) = xp(x) = 1/40 x^2 + 19000x
(b)
P(x) = R(x)-C(x)
= (1/40 x^2 + 19000x)-(15450x + 20000)
= 1/40 x^2 + 3550x - 20000
Now you can find the maximum P(x).
By the way, you never said what x represents. I assumed the price, but maybe I'm wrong.