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\(\text{A shop sells 500 smartphones a week for $\$450$ each. A market survey shows that each decrease of $\$5$ on the price will result in the sale of an additional 10 smartphones per week. What price of the smartphone would result in maximum revenue, in dollars? }\)

 

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 Jul 17, 2019

Best Answer 

 #1
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Suppose the price of the cellphone is reduced to \(450-5x\) dollars; then it will sell \(500x+10\) units, so the revenue will be\(\begin{align*} (450 - 5x)(500 + 10x) &= 5(90 - x) 10(50 + x) \\ &= 50 (90 - x)(50 + x) \\ &= 50 (-x^2 + 40x + 4500), \end{align*}\)

in dollars.

 

Completing the square on \(-x^2 + 40x + 4500,\) we get

\(\begin{align*} 50 (-x^2 + 40x + 4500) &= 50 (-(x - 20)^2 + 400 + 4500) \\ &= 50 (-(x - 20)^2 + 4900) \\ &= -50 (x - 20)^2 + 245000. \end{align*}\)

This is maximized when \(x=20,\) so the optimal price of the smartphone is \(450 - 5(20) = \boxed{350}\) dollars.

.
 Jul 17, 2019
edited by Guest  Jul 17, 2019
edited by Guest  Jul 17, 2019
 #1
avatar
+1
Best Answer

Suppose the price of the cellphone is reduced to \(450-5x\) dollars; then it will sell \(500x+10\) units, so the revenue will be\(\begin{align*} (450 - 5x)(500 + 10x) &= 5(90 - x) 10(50 + x) \\ &= 50 (90 - x)(50 + x) \\ &= 50 (-x^2 + 40x + 4500), \end{align*}\)

in dollars.

 

Completing the square on \(-x^2 + 40x + 4500,\) we get

\(\begin{align*} 50 (-x^2 + 40x + 4500) &= 50 (-(x - 20)^2 + 400 + 4500) \\ &= 50 (-(x - 20)^2 + 4900) \\ &= -50 (x - 20)^2 + 245000. \end{align*}\)

This is maximized when \(x=20,\) so the optimal price of the smartphone is \(450 - 5(20) = \boxed{350}\) dollars.

Guest Jul 17, 2019
edited by Guest  Jul 17, 2019
edited by Guest  Jul 17, 2019

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