Question

a company manufactures and sells x cellphones per week. the weekly price - demand and cost equations are given below.
p = 600 - 0.5x and c(x) = 15,000 + 135x
(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.)
the maximum weekly revenue is $□$. (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.)
the maximum weekly profit is $□$. (round to the nearest cent as needed.)

Explanation:

Step1: Revenue function

Revenue \( R(x)=p\times x=(600 - 0.5x)x=600x-0.5x^{2}\).
To find the maximum of \( R(x)\), we use the formula for the vertex of a parabola \( y = ax^{2}+bx + c\). The \( x\)-coordinate of the vertex is \( x=-\frac{b}{2a}\). For \( R(x)=- 0.5x^{2}+600x\), \( a=-0.5\), \( b = 600\).
\( x=-\frac{600}{2\times(-0.5)}=600\).
Substitute \( x = 600\) into \( p=600 - 0.5x\), \( p=600-0.5\times600 = 300\).
Substitute \( x = 600\) into \( R(x)\), \( R(600)=600\times600-0.5\times600^{2}=180000\).

Step2: Profit function

Profit \( P(x)=R(x)-C(x)=(600x - 0.5x^{2})-(15000 + 135x)=-0.5x^{2}+465x - 15000\).
For the parabola \( y=-0.5x^{2}+465x - 15000\), \( a=-0.5\), \( b = 465\).
The \( x\)-coordinate of the vertex is \( x=-\frac{465}{2\times(-0.5)}=465\).
Substitute \( x = 465\) into \( p=600-0.5x\), \( p=600-0.5\times465=367.5\).
Substitute \( x = 465\) into \( P(x)\), \( P(465)=-0.5\times465^{2}+465\times465-15000\)
\( P(465)=-0.5\times216225 + 216225-15000=93562.5\).

Answer:

(A) The company should produce \(600\) phones each week at a price of \(\$300\). The maximum weekly revenue is \(\$180000\).
(B) The company should produce \(465\) phones each week at a price of \(\$367.5\). The maximum weekly profit is \(\$93562.5\).