Question

a wire 34 cm long is cut into 2 pieces. One piece is bent to form a square and the other is bent to form a rectangle that is twice as wide as it is long. How should the wire be cut in order to minimize the total area of the square and the rectangle?

Answer 1

You're given that the total perimeter of both wire shapes is given by
\[34=4x+2y+2z,\text{ where}\]
x = length of side of square,
y = length of rectangle, and
z = width of rectangle.

You're told that the rectangle's width is twice its length, so you know \(z=2y.\) Substituting, your perimeter equation becomes
\[34=4x+2y+4y\\
34 = 4x + 6y\]
The total area of both pieces is given by
\[A=x^2+yz,\text{ or, using the sub from earlier,}\\
A=x^2+2y^2\]
Find an expression for x in terms of y (or y in terms of x) which you will sub into the area equation. From there, you apply the derivative test and find the y (or x, depending on your expression) that minimizes the total area.

Answer 2

thank you so much! but can you help me solve it? I dont know how to find the expression in terms of x or y and then apply the derivative test. Thank you!

Answer 3

\[4x+6y=34\Rightarrow \color{red}{y=\frac{17-2x}{3}}\]
\[\begin{align*}A&=x^2+2y^2\\
&=x^2+2\left(\color{red}{\frac{17-2x}{3}}\right)^2\\
&=x^2+2\left(\frac{289-68x+4x^2}{9}\right)\\
A&=\frac{17}{9}x^2-\frac{136}{9}x+\frac{578}{9}\end{align*}\]
Now apply the first derivative test, i.e. find the critical points of A (values of x that make A' = 0), then determine over which intervals A is increasing or decreasing.
\[A'=\frac{34}{9}x-\frac{136}{9}\]
Set \(A'=0:\)
\[0=\frac{34}{9}x-\frac{136}{9}\Rightarrow x=4\]
The intervals you thus have to check are \((0,4)\text{ and }(4,\infty)\). This is because x can't be negative or zero (if x = 0, you don't have a square, which means you don't have the two pieces of wire to begin with).

Over the first interval, pick a convenient x-value (such as \(x=1\)), and evaluate the derivative for that value. Sticking with 1, you have \(\displaystyle A'(1)=-\frac{102}{9}\), which tells you that A is decreasing. (Keep in mind that \(A'>0\Rightarrow A\text{ increasing, and }A'<0\Rightarrow A\text{ decreasing.} \))

Over the second interval, pick \(x=5\) and find that \(\displaystyle A'(5)=\frac{34}{9}>0\Rightarrow A\text{ increasing.}\)

The decrease-then-increase indicates a minimum, so the test confirms that a minimum occurs for x = 4.

Answer 4

Since the minimal area occurs for x = 4, you'd cut a piece of wire with length 4*4=16 for the square piece, and leave the rest for the rectangular piece.