I'm having trouble with this problem:
If
$C(x) = 14000 + 500x − 4.8x^2 + 0.004x^3$
is the cost function and
$p(x) = 4100 − 9x$
is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)
I think that in order to find the answer, I have to find the derivatives of both the equations and set them equal to each other. However, I am getting multiple roots and none of the roots are the answers. What should I be looking for if not the roots?
$\endgroup$2 Answers
$\begingroup$The revenue is $x\cdot p(x)$. If you take the derivative of that and set it equal to the derivative of cost, I find a single positive solution.
$\endgroup$ $\begingroup$Let $P(x)$ be the profit function, then $$ \begin{align} P(x)&=x\cdot p(x)-C(x)\\ &=x(4100 − 9x)-(14000 + 500x − 4.8x^2 + 0.004x^3)\\ &=4100x-9x^2-14000 - 500x + 4.8x^2 - 0.004x^3\\ &=-14000+3600x-4.2x^2- 0.004x^3.\\ \end{align} $$ To maximize the profit function, take its first derivative and set equal to zero. $$ \begin{align} \frac{dP}{dx}&=0\\ \frac{d}{dx}(-14000+3600x-4.2x^2- 0.004x^3)&=0\\ 3600-8.4x-0.012x^2&=0.\tag1 \end{align} $$ Multiply $(1)$ by $\dfrac{1000}{12}$, yield $$ \begin{align} 300000-700x-x^2&=0\\ x^2+700x-300000&=0\\ (x-300)(x+1000)&=0. \end{align} $$ Since $x\ge 0$, then the possible solution is $x=300$. The maximum profit is $P(300)$.
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