A company manufactures and sells x cellphones per week. The weekly price-demand and cost equations are given below P=400-.5x and C(x)=20000+130x

What price should the company charge for the phones and how many phones should be produced to maximize the weekly revenue. What is maximum weekly Revenue?

revenue is sales*price

R(x) = x(400-.5x) = 400x - .5x^2
dR/dx = 400 - x
dR/dx = 0 when x = 400
R(400) = 400(400-200) = 80000

Cost doesn't figure into revenue. Now, if you want profit, subtract cost from revenue.

To find the price the company should charge and the quantity of phones to produce in order to maximize weekly revenue, we need to find the derivative of the revenue function and set it equal to zero.

The revenue function can be given by R(x) = x * P(x), where R(x) represents the weekly revenue, x represents the quantity of phones produced, and P(x) represents the price per phone.

Given that the price-demand equation is P(x) = 400 - 0.5x, we can substitute this into the revenue function:

R(x) = x * (400 - 0.5x) = 400x - 0.5x^2

To find the maximum value of R(x), we take the derivative of R(x) with respect to x and set it equal to zero:

R'(x) = 400 - x = 0

Solving for x, we find x = 400.

Therefore, the company should produce 400 phones per week to maximize weekly revenue.

To find the price the company should charge, we substitute x = 400 into the price-demand equation:

P(400) = 400 - 0.5(400) = 400 - 200 = 200

The company should charge a price of $200 per phone to maximize weekly revenue.

To calculate the maximum weekly revenue, we substitute x = 400 into the revenue function:

R(400) = 400(200) - 0.5(400^2) = 80,000 - 80,000 = 0

The maximum weekly revenue is $0. However, it's important to note that this result may not be realistic in a practical business scenario, so the numbers in this example may need to be adjusted accordingly.

To find the price and quantity that will maximize the weekly revenue, we need to determine the point of maximum revenue. The revenue is calculated by multiplying the price by the quantity sold.

First, let's calculate the weekly revenue equation:

Revenue (R) = Price (P) * Quantity (x)

The price equation given is:
P = 400 - 0.5x

Substituting this equation into the revenue equation, we get:

R = (400 - 0.5x) * x
R = 400x - 0.5x^2

To find the maximum revenue, we need to find the maximum point of this quadratic equation. The maximum point of a quadratic equation occurs at its vertex, which lies on the axis of symmetry. The axis of symmetry formula for a quadratic equation of the form ax^2 + bx + c is given by:

x = -b / 2a

In our case, a = -0.5, and b = 400. Substituting these values into the formula, we get:

x = -400 / 2(-0.5)
x = -400 / -1
x = 400

So, the quantity of phones that should be produced to maximize the weekly revenue is 400.

To find the price that should be charged for the phones at this quantity, we can substitute the value of x back into the price equation:

P = 400 - 0.5(400)
P = 400 - 200
P = 200

Therefore, the company should charge a price of 200 for the phones to maximize the weekly revenue.

To calculate the maximum weekly revenue, we substitute the value of x into the revenue equation:

R = (400 - 0.5(400)) * 400
R = (400 - 200) * 400
R = 200 * 400
R = 80,000

Therefore, the maximum weekly revenue is 80,000.