Question

PLEASE ANSWER ASAP

The Pear company sells pPhones. The cost to manufacture x pPhones is
C(x)= -25x^2 + 49000x +21390 dollars (this includes overhead costs and production costs
for each pPhone). If the company sells x pPhones for the maximum price they can fetch, the
revenue function will be R(x)= -29x^2 + 129000x dollars.
How many pPhones should the Pear company produce and sell to maximimze profit? (Remember
that profit=revenue-cost)
x=____

Answer 1

10,000 pPhones.

Explanation:

The cost to manufacture x pPhones is given by the cost function:

`C(x)=-25x^2+49000x+21390`

And the revenue for selling x pPhones is given by the revenue function:

`R(x)=-29x^2+129000x`

We want to determine the number of pPhones the Pear Company should sell to maximize profits.

First, we can find the profit function. Profit is given by:

`P(x)=R(x)-C(x)`

Therefore:

`P(x)=(-29x^2+129000x)-(-25x^2+49000x+21390)`

Simplify:

`P(x)=(-29x^2+129000x)+(25x^2-49000x-21390)`

Add. So, our profit function is:

`P(x)=-4x^2+80000x-21390`

Note that our profit function is a quadratic. The maximum point of a quadratic is always its vertex. So, we can find the vertex of the profit function.

The coordinate of the vertex is given by:

`\displaystyle \Big(-(b)/(2a),f\Big(-(b)/(2a)\Big)\Big)`

In this case, a = -4, b = 80000, and c = - 21390.

Therefore, the point at which profit is maximized is:

`\displaystyle x=-(80000)/(2(-4))=-(80000)/(-8)=10000`

Therefore, in order to maximize profits, the Pear Company should sell exactly 10,000 phones.

Notes:

And the maximum profit will be:

`P(10000)=-4(10000)^2+80000(10000)-21390=\$399,978,610`