The CarryItAll mini-van, a popular vehicle among soccer moms, obeys the demand equation
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?
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.
a) To express the revenue as a function of x (number of vans), we use the formula:
Revenue (R) = Price (p) * Quantity Sold (x)
Given the demand equation p = -(1/40)x + 19,000 - 140x, the price can be substituted into the revenue formula as follows:
R(x) = (-(1/40)x + 19,000 - 140x) * x
= -x/40 + 19,000x - 140x^2
Therefore, the revenue function is R(x) = -x/40 + 19,000x - 140x^2.
b) The profit is calculated by subtracting the cost (C) from the revenue (R):
Profit (P) = Revenue (R) - Cost (C)
Given that the cost function is C(x) = 15,450x + 20,000, we can substitute the revenue function from part (a) into the profit formula:
P(x) = R(x) - C(x)
= (-x/40 + 19,000x - 140x^2) - (15,450x + 20,000)
= -140x^2 + 18,550x - 20,000
Therefore, the profit function is P(x) = -140x^2 + 18,550x - 20,000.
c) To find the value of x that maximizes profit, we can use the critical points method. The maximum occurs at the critical point where the derivative of the profit function is equal to zero.
P'(x) = -280x + 18,550
Setting P'(x) equal to zero and solving for x:
-280x + 18,550 = 0
-280x = -18,550
x = -18,550 / -280
x = 66.25
Since the number of vans (x) cannot be a decimal, we can round it to the nearest whole number. Therefore, the value of x that maximizes profit is 66.
To find the maximum profit, substitute the value of x into the profit function:
P(66) = -140(66)^2 + 18,550(66) - 20,000
Solving this equation will give us the maximum profit.
d) To find the price that should be charged in order to maximize profit, we need to substitute the value of x (66) into the demand equation:
p = -(1/40)(66) + 19,000 - 140(66)
Solving this equation will give us the price that should be charged.
a) Revenue (R) is equal to the product of the price (p) and the quantity sold (x). From the demand equation, we have p = -1/40x + 19,000 - 140x. Therefore, revenue (R) can be expressed as:
R = p * x
R = (-1/40x + 19,000 - 140x) * x
R = -1/40x^2 + 19,000x - 140x^2
b) Profit (P) is equal to revenue (R) minus the cost (C). From part a, we have the revenue (R) equation. From the cost function, C(x) = 15,450x + 20,000, we can express profit (P) as:
P = R - C
P = (-1/40x^2 + 19,000x - 140x^2) - (15,450x + 20,000)
P = -1/40x^2 - 140x^2 + 19,000x - 15,450x - 20,000
c) To find the value of x that maximizes profit, we need to find the derivative of the profit function with respect to x, set it equal to zero, and solve for x. Then, we can substitute that value of x into the profit function to find the maximum profit.
dP/dx = -2/40x - 280x + 19,000 - 15,450
Setting dP/dx = 0:
-2/40x - 280x + 19,000 - 15,450 = 0
-1/20x - 280x = 15,450 - 19,000
-1/20x - 280x = -3550
-1/20x = -3550 + 280x
(279x)/20 = 3550
279x = 20 * 3550
279x = 71000
x = 71000/279
x ≈ 254.48
Substituting x = 254.48 into the profit function to find the maximum profit:
P = -1/40(254.48)^2 - 140(254.48)^2 + 19,000(254.48) - 15,450(254.48) - 20,000
The result will give you the maximum profit.
d) To find the price that should be charged to maximize profit, substitute the value of x (254.48) that maximizes profit into the demand equation p = -1/40x + 19,000 - 140x. This will give you the price that should be charged to maximize profit.