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: 
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 $A(N) > A(0)$.
Let l be the length of wire and n and m be the side lengths of equilateral triangle and that of square respectively.
$$Hence\,\,l = 3n + 4m -------(1) $$

$$Area \,\,A(n)=(\frac{\sqrt{3}}{4})n^2+m^2$$
Using (1) 
$$ A(n)=(\frac{\sqrt{3}}{4})n^2+(\frac{l-3n}{4})^2$$
Differentiate w.r.t n 
$$A'(n)=2.(\frac{\sqrt{3}}{4})n-2\frac{3}{16}(l-3n)$$
$$A'(n)=(\frac{\sqrt{3}}{2}+\frac{9}{8})n-(\frac{3}{8})l$$
Hence Area will be minimum when 
$$n=(\frac{(\frac{3}{8})}{\frac{\sqrt{3}}{2}+\frac{9}{8}})l$$
$$n=(\frac{3}{4\sqrt{3}+9})l$$
Hence maximum will be either when n=0 or m=0.