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a company manufactures and sells x cellphones per week. the weekly pric…

Question

a company manufactures and sells x cellphones per week. the weekly price - demand and cost equations are given below.
p = 500 - 0.1x and c(x)=25,000 + 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.)
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: Find the revenue function

Revenue \( R(x)=p\times x=(500 - 0.1x)x = 500x-0.1x^{2}\)

Step2: Find the derivative of the revenue function

\(R^{\prime}(x)=\frac{d}{dx}(500x - 0.1x^{2})=500 - 0.2x\)

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

\(500-0.2x = 0\)
\(0.2x=500\)
\(x = 2500\)

Step4: Find the price \(p\)

Substitute \(x = 2500\) into \(p = 500-0.1x\), \(p=500-0.1\times2500=250\)

Step5: Find the maximum revenue

Substitute \(x = 2500\) into \(R(x)\), \(R(2500)=500\times2500-0.1\times(2500)^{2}\)
\(R(2500)=1250000 - 625000=625000\)

Step6: Find the profit function

Profit \(P(x)=R(x)-C(x)=(500x - 0.1x^{2})-(25000 + 140x)=- 0.1x^{2}+360x - 25000\)

Step7: Find the derivative of the profit function

\(P^{\prime}(x)=\frac{d}{dx}(-0.1x^{2}+360x - 25000)=-0.2x + 360\)

Step8: Set the derivative equal to zero to find critical points

\(-0.2x+360 = 0\)
\(0.2x=360\)
\(x = 1800\)

Step9: Find the price \(p\)

Substitute \(x = 1800\) into \(p = 500-0.1x\), \(p=500-0.1\times1800 = 320\)

Step10: Find the maximum profit

Substitute \(x = 1800\) into \(P(x)\), \(P(1800)=-0.1\times(1800)^{2}+360\times1800-25000\)
\(P(1800)=-324000+648000 - 25000=299000\)

Answer:

(A) The company should produce \(2500\) phones each week at a price of \(\$250\). The maximum weekly revenue is \(\$625000\).
(B) The company should produce \(1800\) phones each week at a price of \(\$320\). The maximum weekly profit is \(\$299000\).