Question

The monthly profit from the sale of a product is given by P = 18x − 0.1x² − 100 dollars.

(a) What level of production maximizes profit?
(b) What is the maximum possible profit?

Answer 1

Solution :-

Given data :

The monthly profit from the sale of a product is given by P = 18x − 0.1x² − 100 dollars.

(a) What level of production maximizes profit?

Here consider differentiation of P with respect to x, we get

$\frac{dP}{dx}=\frac{d}{dx}(18x-0.1x^2-100)$

$\frac{d^2P}{dx^2}=\frac{d}{dx}(18x)-\frac{d}{dx}(0.1x^2) -\frac{d}{dx}(100)$

$\frac{dP}{dx}=18-0.1*2x$

$\frac{dP}{dx}=18-0.2x$

Again consider differentiation with respect to x, we get

$\frac{d^2P}{dx^2}=\frac{d}{dx}\left \{\frac{dP}{dx} \right \}$

$\frac{d^2P}{dx^2}=\frac{d}{dx}\left \{18-0.2x \right \}$

$\frac{d^2P}{dx^2}=\frac{d}{dx}(18)-\frac{d}{dx}(0.2x)$

$\frac{d^2P}{dx^2}=-0.2$ < 0

Here,P will be maximum at critical point. i.e  $\frac{dP}{dx}=0$

$\Rightarrow$ 18 - 0.2 x =0

$\Rightarrow$ 18 = 0.2 x

$\Rightarrow$  $x=\frac{18}{0.2}$

$\Rightarrow$ x = 90

$\Rightarrow$Therefore, level of production maximizes profit ,x = 90.

(b) What is the maximum possible profit?

$P_{max}=18x-0.1x^2-100$

$P_{max}=18(90)-0.1(90)^2-100$

$P_{max}=1620-810-100$

$P_{max}=710$

$\Rightarrow$The maximum possible profit, $P_{max}=710$