Question: You have a length of wire and you want to find the maximum and minimum area that can be created if you cut the wire into two, and bend it to form a square and an equilateral triangle. 

I know the answer will be that the maximum will be when all the wire is used for the square, and the minimum will be when all the wire is used for the triangle but I don’t know how to prove that. 

I’ve showed my work in the attached picture. I tried using a length of 100 to help solve but that didn’t really improve my understanding at all. 

[![enter image description here](https://i.stack.imgur.com/5gCxx.jpg)](https://i.stack.imgur.com/5gCxx.jpg)

(<https://i.stack.imgur.com/5gCxx.jpg>)

Answer: 
Assume the wire is $L$ long and you cut it to $xL$ for the triangle and $(1-x)L$ for the square. Then, the side of the equilateral triangle must be $t=xL/3$ and of the square will be $s=(1-x)L/4$.

Based on this, the area of the triangle will be $t^2\sqrt{3}/4$ and of the square will be $s^2$, so the total area now can be found as
$$
A(x) = x^2L^2 \frac{\sqrt{3}}{4 \cdot 9} + \frac{(1-x)^2L^2}{16},
$$
and now you are in a familiar world of Calculus I with optimization over $0 \le x \le 1$.

Can you finish it?
Use $N$ for the total length (as a constant), and $x$ (variable) as the length of the side of a square. Then, the length of the sides of the triangle will be $\frac{N - x}{3}$.

Then you can use $\frac{N - x}{3}$ as the base for your triangle, and $(\frac{1}{4}x)^2$ as the area of the square. Then you create a function $A(x)$ that computes the total of the two areas. 

This simplifies to a quadratic opening up. However, because it is a geometric figure, it has specific valid values since we can't have negative side lengths. The interval it is valid for is $x = 0$ to $x = N$. Since it is a quadratic, it only has a local minimum. Therefore, the maximas are at the edges of the graph. Therefore, the maximum *has* to be at either $x = 0$ or $x = N$. So, you have to show whether $A(0) > A(N)$ or $A(N) > A(0)$.