can someone show me how to solve this problem.
It determines that in order to sell x smartphones, the price per product must be p=280-0.4x. It also determine that the total cost of producing x smartphones is given by C(x)=5000+0.6x.
a. Find the total revenue, R(x).
b. Find the total profit, P(x).
c. How many smartphones must the company sell in order to maximize profit?
d. What is the maximum profit?
e. What is th price per smartphone must be charged in order to maximize profit?
can u please show me the steps to solve this and the final ans. x need to calculate.
R(x) = p(x)*x
P(x) = R(x)-C(x)
P(x) = x(280-.4x) - (5000+.6x)
= -.4x^2 + 279.4x - 5000
This is a parabola, with maximum at x = -279.4/-.8 = 349.25
plug that x into the functions to get profit and price.
To solve this problem, we need to follow a step-by-step approach. Let's go through each part and find the solutions:
a. Total Revenue, R(x)
The total revenue is calculated by multiplying the price per product (p) by the number of products sold (x). In this case, the price per product is given by p = 280 - 0.4x. Therefore, the total revenue can be calculated as:
R(x) = p * x = (280 - 0.4x) * x = 280x - 0.4x^2
b. Total Profit, P(x)
Profit is calculated by subtracting the total cost (C(x)) from the total revenue (R(x)). The total cost is given by C(x) = 5000 + 0.6x. Therefore, the total profit can be calculated as:
P(x) = R(x) - C(x) = (280x - 0.4x^2) - (5000 + 0.6x) = 280x - 0.4x^2 - 5000 - 0.6x = -0.4x^2 + 279.4x - 5000
c. Maximizing Profit
To find the number of smartphones that maximize profit, we need to find the x-value that corresponds to the maximum value of P(x). This can be done by finding the derivative of P(x) with respect to x, setting it equal to zero, and solving for x.
dP(x)/dx = -0.8x + 279.4 = 0
-0.8x = -279.4
x = -279.4 / -0.8
x = 349.25
Since the number of smartphones sold cannot be a fraction, the company must sell 349 smartphones to maximize profit.
d. Maximum Profit
To find the maximum profit, substitute the value of x into the profit equation P(x) = -0.4x^2 + 279.4x - 5000:
P(349) = -0.4(349)^2 + 279.4(349) - 5000
P(349) = -0.4(121,801) + 97,305.6 - 5000
P(349) = -48,720.4 + 97,305.6 - 5000
P(349) = 43,585.2
Therefore, the maximum profit is $43,585.20.
e. Price per Smartphone for Maximum Profit
To find the price per smartphone that maximizes profit, substitute the value of x into the price equation p = 280 - 0.4x:
p = 280 - 0.4(349)
p = 280 - 139.6
p = 140.4
Therefore, the price per smartphone must be set at $140.40 to maximize profit.
To summarize:
a. Total revenue, R(x) = 280x - 0.4x^2
b. Total profit, P(x) = -0.4x^2 + 279.4x - 5000
c. The company must sell 349 smartphones to maximize profit.
d. The maximum profit is $43,585.20.
e. The price per smartphone for maximum profit is $140.40.