Hello, stealthmaths!

A piece of wire of total length 12 m is cut into two pieces.

One piece of wire is bent into a rectangle of sides $\displaystyle x$ m and $\displaystyle 3x$ m

and the other is bent to form the boundary of a square.

Show that the total area enclosed by the rectangle and the square

is given by: . $\displaystyle A \;=\;7x^2 - 12x + 9$

Find the value of $\displaystyle x$ if the total area enclosed is a minimum.

Verify that your stationary value is a minimum.

The 12-meter wire is cut into two pieces of length $\displaystyle a$ and $\displaystyle 12-a$.

Code:

: - - - - - 12 - - - - - :
*---------*---------------*
: - a - : - - 12-a - - :

The left piece of wire is bent into a $\displaystyle x\text{-by-}3x\text{ rectangle.}$ Code:

3x
* - - - - - *
| |
x | | x
| |
* - - - - - *
3x

The area of this rectangle is: .$\displaystyle 3x\cdot x \:=\:\boxed{3x^2}$

We have: .$\displaystyle x + 3x + x + 3x \:=\:a\quad\Rightarrow\quad 8x \:=\:a$

The right piece of wire has length: .$\displaystyle 12-8x$

It is bent into a square. Code:

s
* - - - *
| |
s | | s
| |
* - - - *
s

We have: .$\displaystyle 4s = 12 - 8x \quad\Rightarrow\quad s \:=\:3 - 2x$

The area of this square is: .$\displaystyle \boxed{(3-2x)^2}$

Therefore, the total area is: .$\displaystyle A \;=\;3x^2 + (9 - 6x + 4x^2) \;=\;7x^2 - 6x + 9 $