Could you please give me any advices how to solve this exercise? Would be grateful for any help:)
A firm's market demand function is $AR = 150-3x$ , and its total cost function is $TC= 0.1x^3 – 3x^2 + 30x$. Find the level of output to maximize profit.
Best Answer
The revenue function of the firm is $R(x)=AR\cdot x=p(x)\cdot x=150x-3x^2$
And the total cost function is $C(x)=TC=0.1x^3 - 3x^2 + 30x$
Therfore the profit function is $P(x)=R(x)-C(x)=150x-3x^2-(0.1x^3-3x^2+30x)=120x-0.1x^3$
To get an extreme point, $x^*$, we have to calculate the derivative of $P(x)$ and set it equal to zero:
$P'(x)=120-0.3x^2=0 \qquad $
Adding $0.3x^2$ on both sides.
$120=0.3x^2$
Dividing both sides by $0.3=\frac3{10}$
$400=x^2$
$x_{1/2}=\pm 20$
Quantities are always positive. Thus $x^*=20$
For the last you have to check if this point is a (relative) maximum.
$P''(x^*)<0 \Rightarrow x^* \texttt{is a maximum}$