Question

a company manufactures and sells x cellphones per week. the weekly price - demand and cost equations are given below.
p = 400 - 0.5x and c(x)=20,000 + 140x
(a) what price should the company charge for the phones, and how many phones should be produced to maximize the weekly revenue? what is the maximum weekly revenue?
the company should produce □ phones each week at a price of $ □.
(round to the nearest cent as needed.)

Explanation:

Step1: Find the revenue function

Revenue \(R(x)=p\times x=(400 - 0.5x)x = 400x-0.5x^{2}\).

Step2: Find the derivative of the revenue function

Using the power rule \((x^n)^\prime=nx^{n - 1}\), \(R^\prime(x)=\frac{d}{dx}(400x-0.5x^{2})=400 - x\).

Step3: Set the derivative equal to zero to find critical points

Set \(R^\prime(x)=0\), so \(400 - x = 0\), which gives \(x = 400\).

Step4: Check the second - derivative

\(R^{\prime\prime}(x)=\frac{d}{dx}(400 - x)=-1<0\). Since the second - derivative is negative, \(x = 400\) is a maximum.

Step5: Find the price

Substitute \(x = 400\) into the price - demand equation \(p = 400-0.5x\). Then \(p=400-0.5\times400=400 - 200=\$200\).

Step6: Find the maximum revenue

Substitute \(x = 400\) into the revenue function \(R(x)=400x-0.5x^{2}\). \(R(400)=400\times400-0.5\times400^{2}=160000-0.5\times160000=80000\).

Answer:

The company should produce \(400\) phones each week at a price of \(\$200\). The maximum weekly revenue is \(\$80000\).