0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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? $\endgroup$
    – David
    May 21, 2014 at 1:54

4 Answers 4

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.)

$\endgroup$
0
$\begingroup$

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 $x^2+(10-x)^2 = 2(x-5)^2+50$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.