We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
57
1
avatar

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

 

Thanks!

 Jul 17, 2019

Best Answer 

 #1
avatar
+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.

.
 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

12 Online Users

avatar