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

Thanks!

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

edited by Guest Jul 17, 2019

#1**+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

edited by Guest Jul 17, 2019