Sony Ent. is marketing a new smartphone. 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^2 .
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 price per smartphone must be charged in order to maximize profit?
To find the total revenue, R(x), we need to multiply the price per product, p, by the number of smartphones sold, x:
R(x) = p * x
Given that p = 280 - 0.4x, we substitute this value into the equation to get:
R(x) = (280 - 0.4x) * x
Simplifying, we expand the expression:
R(x) = 280x - 0.4x^2
So, the total revenue function is R(x) = 280x - 0.4x^2.
To find the total profit, P(x), we subtract the total cost, c(x), from the total revenue, R(x):
P(x) = R(x) - c(x)
Given that R(x) = 280x - 0.4x^2 and c(x) = 5000 + 0.6x^2, we substitute these values:
P(x) = (280x - 0.4x^2) - (5000 + 0.6x^2)
Simplifying, we combine like terms:
P(x) = 280x - 0.4x^2 - 5000 - 0.6x^2
P(x) = -1x^2 + 280x - 5000
So, the total profit function is P(x) = -x^2 + 280x - 5000.
To find the number of smartphones the company must sell in order to maximize profit, we can use the vertex formula for a quadratic function:
x = -b / (2a)
In this case, a = -1 and b = 280. Substituting these values into the formula:
x = -280 / (2 * -1)
x = -280 / -2
x = 140
The company must sell 140 smartphones to maximize profit.
To find the maximum profit, we substitute the value of x = 140 into the profit function P(x):
P(140) = -(140)^2 + 280(140) - 5000
P(140) = -19600 + 39200 - 5000
P(140) = 14600
The maximum profit is $14,600.
To find the price per smartphone that maximizes profit, we substitute the value of x = 140 into the price function p:
p = 280 - 0.4x
p = 280 - 0.4(140)
p = 280 - 56
p = 224
The price per smartphone that maximizes profit is $224.