# The Sam H. Aunted Co. has inverse demand curve p = 172 - 8x and cost function C(x) = 28 + 0.6x +...

## Question:

The Sam H. Aunted Co. has inverse demand curve {eq}p = 172 - 8x {/eq} and cost function {eq}C(x) = 28 + 0.6x + 13x^2 {/eq}.

Find the quantity {eq}x {/eq} that maximizes profit.

## Application of Derivative:

The above problem concerns the topic of the application of derivative as maxima and minima. A function attains its maximum or minimum value at a critical point. The critical point is obtained by equating the first derivative of the function to zero.

Revenue function is the product of demand function and the quantity produced. Profit function is the difference between revenue function and the cost function

Given inverse demand function {eq}p = 172 - 8x {/eq} and cost function {eq}C(x) = 28 + 0.6x + 13x^2 {/eq}, where {eq}x {/eq} is the quantity produced.

First, revenue function is the product of demand function and the number of quantity produced. Revenue functon is given as

{eq}\begin{align} R(x) &= p(x) \cdot x \\ R(x) &= (172-8x )x \\ R(x) &= 172x -8x^2 \end{align} {/eq}

Now the profit is the difference between revenue generated and the cost function. It is given as

{eq}\displaystyle \begin{align} P(x) &= R(x)- C(x) \\ P(x) &= 172x-8x^2-( 28 + 0.6x + 13x^2) \\ P(x) &= -21x^2+ 171.4 x-28 \end{align} {/eq}

Differentiating above concerning {eq}x {/eq} and equation it to zero, we obtain

{eq}\displaystyle \begin{align} P'(x) &= -42x+ 171.4 =0 \\ 42x &= 171.4 \\ x &= 4.08 \end{align} {/eq}

Now {eq}\displaystyle \frac{\mathrm{d^2} P(x)}{\mathrm{d} x^2} =-42<0 {/eq}, so profit is maximum for {eq}x=4.08 \approx 4 {/eq}

So quantity produced must be {eq}{\color{Blue} {4}} {/eq}, to maximize the profit. 