So I have a question about a farely classical example of an optimization problem. It concerns optimizing the volume of a box. I know the general method of solving this using the derivative but I'm wondering if there are any other, more elementary strategies of optimizing the volume? That is, strategies which to not involve the concept of the derivative. Thanks in advance! The optimization problem in question is seen here below.
The question goes as follows: You have been given the task of folding up a square-shaped sheet of metal with side $a=54$ cm to an open box (see figure). This is done by first chopping of four squares with side $x$ on the corners of the metal sheet. After this we fold up the sides and weld them togehter into a box.
- Write an expression for the volume of the box as a function of $x$.
- What is the domain of the volume function (what are the allowed values of $x$?)
- What dimensions should the box have in order to yield the maximum volume? What is the maximum volume?
I've found one method which is called Fermats method to solve extreme value problems.
One may observe that in the vicinity of an extreme value of a function, the value of the function is constant. That is, $V(x)=V(x+\varepsilon )$ where $\varepsilon>0$ is some small number ($\varepsilon\rightarrow0$, really). Solving the above equation with respect to $x$ yields the following rather messy equation $$\varepsilon (4\varepsilon ^{2}+12\varepsilon x-216\varepsilon +12x^{2}-432x+2916\varepsilon)=0.$$ Now, one may do something a little bit "illegal" which is to divide by $\varepsilon$ which yields $$4\varepsilon ^{2}+12\varepsilon x-216\varepsilon +12x^{2}-432x+2916\varepsilon=0.$$ If we let $\varepsilon =0$, this will yield an equation involving a second-degree polynomial which corresponds to $V'(x)$! $$12x^{2}-432x+2916=0\Leftrightarrow x=27~~\text{or}~~x=9$$. $V(x)$ is only defined for $0<x<27$ so we may only choose $x=9$.