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 400 phones each week at a price of $200
(round to the nearest cent as needed.)
the maximum weekly revenue is $80000 (round to the nearest cent as needed.)
(b) what price should the company charge for the phones, and how many phones should be produced to maximize the weekly profit? what is the maximum weekly profit?
the company should produce □ phones each week at a price of $□
(round to the nearest cent as needed.)

Explanation:

Step1: Find the revenue function

The revenue function \(R(x)\) is given by \(R(x)=x\times p\). Substituting \(p = 400-0.5x\) into the revenue formula, we get \(R(x)=x(400 - 0.5x)=400x-0.5x^{2}\).

Step2: Find the profit function

The profit function \(P(x)\) is given by \(P(x)=R(x)-C(x)\). Substituting \(R(x)=400x - 0.5x^{2}\) and \(C(x)=20000 + 140x\) into the profit formula, we have \(P(x)=(400x-0.5x^{2})-(20000 + 140x)=- 0.5x^{2}+260x - 20000\).

Step3: Find the vertex of the profit function

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-0.5\), \(b = 260\), \(c=-20000\)), the \(x\) - coordinate of the vertex is \(x=-\frac{b}{2a}\).
Substituting \(a=-0.5\) and \(b = 260\) into the formula \(x=-\frac{b}{2a}\), we get \(x=-\frac{260}{2\times(-0.5)}=260\).

Step4: Find the price

Substitute \(x = 260\) into the price - demand equation \(p = 400-0.5x\). Then \(p=400-0.5\times260=400 - 130=270\).

Step5: Find the maximum profit

Substitute \(x = 260\) into the profit function \(P(x)=-0.5x^{2}+260x - 20000\).
\(P(260)=-0.5\times(260)^{2}+260\times260-20000\)
\(=-0.5\times67600 + 67600-20000\)
\(=-33800+67600 - 20000\)
\(=13800\)

Answer:

The company should produce \(260\) phones each week at a price of \(\$270\). The maximum weekly profit is \(\$13800\).