Science:Math Exam Resources/Courses/MATH104/December 2010/Question 04/Solution 1

Let $A(q)$ denote the average cost per kilogram of producing $q$ kilograms of this chemical,

$A(q)=\frac{C(q)}{q}=3q^{1/3}+50+\frac{10000}{q}.$

The domain of interest is $(0,\mathrm{\infty })$ because the company cannot produce negative kilograms of their product. In order to find the absolute minimum of the average cost function, we begin by looking for its critical points. The derivative of $A$ is

$A^{\prime }(q)=q^{-2/3}-\frac{10000}{q^2}=\frac{1}{q^2}(q^{4/3}-10000)$

This exists everywhere in our domain and is equal to zero when $q^{4/3}-10000=0$, that is, when

$q^{\frac{4}{3}}=10000=10^4.$

Since $q\ge 0$, we can take the positive fourth root of both sides to obtain

$q^{\frac{1}{3}}=10.$

Hence,

$q=10^3=1000$

which is our final answer.

Note:

It is easy to check that this is indeed an absolute minimum by noting that $A^{\prime }(q)<0$ when $q<1000$ and $A^{\prime }(q)>0$ when $q>1000$. Since $A(q)$ is decreasing when $q<1000$ and increasing when $q>1000$, $A(q)$ must have an absolute minimum at $q=1000.$

Hence, the company should produce 1000 kg of chemical in order to minimize its average cost.