# Suppose that when a cell phone store charges 17$ for a car charge, it sells 30 units. When it...

## Question:

Suppose that when a cell phone store charges 17$ for a car charge, it sells 30 units. When it drops the price to 16$ it sells 33 units. Assume that the demand is a linear function of price.

a. If each phone charger costs 1$ to make, what price should the store charge to maximize its profit?

b. If x is the number of times the price is reduced by one dollar, find a function for total profit with respect to x. A negative value for x will mean the price is increased.

## Maximize Profit:

Take profit function as {eq}f(x)=(17-x)(30+3x)-(30+3x) {/eq}

Find {eq}f'(x) {/eq} and put {eq}f'(x)=0 {/eq} to find the critical point.

Find {eq}f''(x) {/eq} to check if that critical point is the point of maxima or minima.

{eq}x=a {/eq} is a point of maxima if {eq}f''(a)<0 {/eq} and {eq}x=a {/eq} is a point of minima if {eq}f''(a)>0 {/eq}

## Answer and Explanation:

As a price that a cell phone charges drops from {eq}17\$ {/eq} to {eq}16\$ {/eq}, price reduced by {eq}17\$-16\$=1\$ {/eq}

Also, the number of units sold increased from 30 units to 33 units.

Profit function:

{eq}f(x)=(17-x)(30+3x)-(30+3x) {/eq}

On differentiating with respect to x, we get

{eq}f'(x)=(17-x)'(30+3x)+(17-x)(30+3x)'-(30+3x)'\\ =-1(30+3x)+3(17-x)-3\\ =-30-3x+51-3x-3\\ =-6x+18 {/eq}

To find: critical points

{eq}f'(x)=0\Rightarrow -6x+18=0\Rightarrow x=3 {/eq}

On differentiating {eq}f'(x) {/eq} with respect to x, we get

{eq}f''(x)=-6\Rightarrow f''(3)=-6<0 {/eq}

So, {eq}x=3 {/eq} is a point of maxima.

Therefore, the store should charge {eq}3\$ {/eq} to maximize profit.

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from General Studies Math: Help & Review

Chapter 5 / Lesson 2