1. Quadratic Functions- Maximum and Minimum Problems

This problem is from the Addison-Wesley Mathematics 11 Text Book (Western Canada Edition)

Section 2.5 - Maximum and Minimum problems

pg. 131 Problem #16

A 30-cm piece of wire is cut in two. One piece is bent into the shape of a square. The other piece is bent into the shape of a rectangle with a length-to-width ratio of 2:1. The sum of the areas of the square and rectangle is a minimum. What are the lengths of the two pieces?

I have tried to solve most of the problem already; I just don't know how to end it. The back of the text says that the answer is 14.1cm, 15.9 cm but it could always be wrong. I would really appreciate if someone could help me Here is my attempt at it: (I’m sorry if it’s messy, I tried to do the best I could on the computer.)

Piece #1 + Piece #2 = 30cm
Piece #1 = x
Piece #2 = 30 - x

Area of the square = 1/4x2

Area of the rectangle = 2w2

6w = 30-x
w = (30-x)
6
w = 5 - (x/6)

Area of the rectangle = 2w2
= 2(5 - (x/6))2
= (10 – (x/3))2

Sum of the areas = 1/4x2 + (10 – (x/3))2

I tired this equation in my graphing calculator but came up with a minimum vertex of x=9.23 and y=69.23, which is impossible because y must be less than 30.

Thank you!

-Olivia

2. Hello, Olivia!

You made a vailiant effort, but your set-up is off . . .

A 30-cm piece of wire is cut in two.
One piece is bent into the shape of a square.
The other piece is bent into the shape of a rectangle with a length-to-width ratio of 2:1.
The sum of the areas of the square and rectangle is a minimum.
What are the lengths of the two pieces?
Let $\displaystyle x$ = length of wire for the rectangle.
Then $\displaystyle 30-x$ = length of wire for the square.

The rectangle looks like this:
Code:
             2k
* - - - - - - - *
|               |
k |               | k
|               |
* - - - - - - - *
2k
The perimeter of the rectangle is: .$\displaystyle 2k + k + 2k + k \:=\:6k$
Its wire is $\displaystyle x$ cm long: .$\displaystyle 6k = x\quad\Rightarrow\quad k = \frac{x}{6}$ . [1]

The area of the rectangle is: .$\displaystyle A_r \:=\:(2k)(k) \:=\:2k^2$ . [2]

Substitute [1] into [2]: .$\displaystyle A_r \;=\;2\left(\frac{x}{6}\right)^2\quad\Rightarrow\q uad\boxed{A_r\:=\:\frac{x^2}{18}}$

The square looks like this:
Code:
      * - - - - - - *
|             |
|             |
|             | ¼(30-x)
|             |
|             |
* - - - - - - *
¼(30-x)
The perimeter of the square is: $\displaystyle 30 - x$ cm,
. . so its side is: $\displaystyle \frac{1}{4}(30 - x)$

The area of the square is: .$\displaystyle A_s \;=\;\left(\frac{30-x}{4}\right)^2\quad\Rightarrow\quad\boxed{A_s\;=\; \frac{x^2 - 60x + 900}{16}}$

Hence, the total area is: .$\displaystyle A \;=\;\frac{x^2}{18} + \frac{x^2 - 60x + 900}{16} \;=\;\frac{1}{144}(17x^2 - 540x + 8100)$

And that is the function we must minimize . . .

3. Thank you

Thank you very much, I really do appreciate it