Question

The pear company sells pphones. The cost to manufacture x pPhones is C(x)= -21x^2+52000x+21087 dollars (this includes overhead costs and production costs for each pphones). If the company sells x pphones for the maximum price they can fetch, the revenue function will be R(x)= -27x2+19600x dollars. How many pphones should the pear company produce and sell to maximize profit?

Answer 1

Answer:

2,700 phones should the pear company produce and sell to maximize profit.

Step-by-step explanation:

The cost to manufacture x phones :

$C(x)= -21x^2+52,000x+21,087$

The revenue function will be :

$R(x)= -27x^2+19,600x$

The profit = P(x) = R(x) - C(x)

$P(x)=(-27x^2+19,600x)-(-21x^2+52,000x+21,087)$

$P(x)=-6x^2+32,400x-21,087$

Differentiating P(x) with respect to dx;

$\frac{d(P(x))}{dx}=\frac{d(-6x^2+32,400x-21,087)}{dx}$

$\frac{d(P(x))}{dx}=-12x+32,400$

$\frac{d(P(x))}{dx}=0$

$0=-12x+32,400$

$12x=32,400$

x = 2,700

Differentiating $\frac{d(P(x))}{dx}$ with respect to dx again;

$\frac{d^2(P(x))}{dx^2}=-12$ (maxima)

2,700 phones should the pear company produce and sell to maximize profit.