Answer:
2,700 phones should the pear company produce and sell to maximize profit.
Step-by-step explanation:
The cost to manufacture x phones :
[tex]C(x)= -21x^2+52,000x+21,087[/tex]
The revenue function will be :
[tex]R(x)= -27x^2+19,600x[/tex]
The profit = P(x) = R(x) - C(x)
[tex]P(x)=(-27x^2+19,600x)-(-21x^2+52,000x+21,087)[/tex]
[tex]P(x)=-6x^2+32,400x-21,087[/tex]
Differentiating P(x) with respect to dx;
[tex]\frac{d(P(x))}{dx}=\frac{d(-6x^2+32,400x-21,087)}{dx}[/tex]
[tex]\frac{d(P(x))}{dx}=-12x+32,400[/tex]
[tex]\frac{d(P(x))}{dx}=0[/tex]
[tex]0=-12x+32,400[/tex]
[tex]12x=32,400[/tex]
x = 2,700
Differentiating [tex]\frac{d(P(x))}{dx}[/tex] with respect to dx again;
[tex]\frac{d^2(P(x))}{dx^2}=-12[/tex] (maxima)
2,700 phones should the pear company produce and sell to maximize profit.
