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

Thanks!

Jul 17, 2019

#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.

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

#1
+1

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