Optimization Question, Finding Maximum and Minimum Values of $30x^2 + 480/x$

Question

The question I have is regarding the construction of a cuboid tank with length and breath $x$ and height $h$ and volume $4$ cubic metres. For the construction of the cuboid, $£15$ per square metre is required for the construction of the top and bottom, and for the four vertical sides $£30$ per square metre. Now, the formula $C=30x^2 + 480/x$ gives the total cost for the construction of the cuboid tank. However, I have to find the values of $x$ (non-negative values of course, since we are dealing with a shape) that give the minimum and maximum cost for the construction of the cuboid, So, I tried to derive the function to locate stationary points but the derived function is a function with asymptotes which means there is no maxm. or min. cost. So, I was wondering if the question is wrong or I am missing something?

Answer 1

we have $$C'(x)=60x-\frac{480}{x^2}=0$$ if $$60x^3=480$$ can you solve this?

Answer 2

60x−480x−2 is the derivative. Put it equal to zero and multiply both sides by x^2. This gives you 60x^3=480. Solve for x

Answer 3

The expression $C'(x) = 0$, where $C'(x)$ is the first derivative of $C(x)$, will give you the maxima and minima of the function $C(x)$. To determine whether or not the extrema is a maximum or a minimum, take the second derivative of $C(x)$, $C''(x)$, at those points $x$ such that $C'(x) = 0$. If $C''(x) > 0$ for those points, then you have a minimum. If $C''(x) < 0$ for those points, you have a maximum. If $C''(x) = 0$, then this result is inconclusive.

Answer 4

You can solve this without differentiation, by using AM-GM inequality.

$$ 30x^2+\frac{480}{x} =30x^2+\frac{240}{x}+\frac{240}{x} \geq (30x^2 \cdot \frac{240}{x} \cdot \frac{240}{x})^{1/3} =1728000^{1/3}=120$$.

This minimum is achieved when $30x^2=\frac{240}{x}$, which is easy to solve.