A piece of wire $20$ metres long is cut into $2$ pieces and each piece is bent to form a square. Determine the length of the two pieces so that the sum of the areas of the two squares is a minimum.
We have to solve it using the vertex method. however i dont know where to begin with. i dont know how to setup the equation. please help. the answer is $10$ metres each.
Hint: Let $x$, and $20-x$ be the length of the $2$ pieces of wire. The sum of areas of the $2$ squares is: $\dfrac{x^2}{16} + \dfrac{(20-x)^2}{16} = f(x) $. Can you simplify it to a quadratic function and use the "vertex" technique?