# max and minimum qudratic function problem

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.

• What is "Vertex Method"? – Babai Nov 24 '14 at 7:54
• okay maybe i am not stating it right. but its the x=-b/2a method."The x coordinate of the vertex can be found by x=-b/2a. substituting the x value of the vertex into the equation of the function yields the y value of the vertex." – sniper007 Nov 24 '14 at 8:13

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?