# Grade 10: Maxima and Minima (Application Question)

This is my Question: A piece of wire 40cm long is to be cut into two pieces which are each bent into the shape of a square. Find the length of each piece of wire if the sum of the areas of the squares is least.

I don't know what should i try and I don't know how to do it.

• Start by letting the length of one piece be $x$. Then what is the length of the other? What are the side lengths of the two squares? What is their area? What is the total area? May 21, 2014 at 1:54

Since you used the tag 'algebra-precalculus', I'm assuming you aren't allowed to differentiate. In that case, let the length of the 2 parts be $x$ and $40-x$. Then the areas of the 2 squares are $\frac{x^2}{16}$ and $\frac{(40-x)^2}{16}$. Then you just add them and complete the square, that is, add or subtract some constant so that the quadratic expression can be written as $$a(x-s)^2+m$$, where $a$ is positive and non-zero and $0\leq s \leq 40$. In that case, $m$ is the minimal sun of their area.
Say the side legnths are $s_1, s_2$. Then, one equation is $4 s_1 + 4 s_2 = 40$ (by the length of the wire).
The area enclosed is $s_1^2+s_2^2$. Solve the equation you got for the length of the wire for either $s_1$ or $s_2$, substitute it into the expression for the area enclosed. You will get a quadratic, which you can minimize for side lengths between $0$ and $10$ (so there is enough wire, and side lengths can't be negative since $s_1 \geq 0, s_2 \geq 0$). The minimum will occur at either the vertex of the parabola defined at the quadratic, at side = 0 or side = 10. Then compute the other side, done.
Suppose you cut the wire into 2 pieces, one of length $x$ and the other of length $40-x$. Forming them into squares, you get one square $x/4$ on a side with area $(x/4)^2$, the other $(40-x)/4$ on a side with area $((40-x)/4)^2$. So the total area of the two squares is given by $A(x)=(x/4)^2+((40-x)/4)^2$. Multiply out and simplify this expression for $A(x)$ -- it's quadratic, and the leading coefficient is positive. What does this tell you about the graph of $A(x)$, and in particular the vertex of the parabola? You can use this to find the minimum area, and the associated value of $x$. (Just as a check: the associated value of $x$ is 20.)
Hint: Let the side of one of the squares be $x$. Then the side of the other square is $10-x$, and the area you want minimised is $x^2+(10-x)^2 = 2(x-5)^2+50$.