# Math Help - calculus for areas

1. ## calculus for areas

Hi:
I have this question that follows. It is in a differentiation section of the book. It asks to find area and value of x. But I thought when I differentiate it is to find the gradient of a tangent to a curve. What don't I know here?

The question is as follows:
A piece of wire of total length 12m is cut into two pieces. One piece of wite is bent into a rectangle of sides xm and 3xm 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:

Find the value of x if the total area enclosed is a minimum. Verify that your stationary value is a minimum.

Thank you

2. Right well lets first find perimeters.

for the rectangle its $x + 3x + x + 3x = 8x$.

So for the square it's just whatever is left which is $12m-8x$.

So area of the rectangle is width times length which is $3x^2$.

Each side of the square has length $\frac{12m - 8x}{4} = 3m - 2x$.

Hence area of the square is...

$(3m - 2x)^2 = 9m^2 - 12x + 4x^2$

Hence total area is...

$4x^2 + 3x^2 - 12x + 9m^2 = 7x^2 - 12x + 9m^2$.

To find the minimum find the derivative and set it to be 0.

To show it is a minimum find the second derivative and check if it is greater than 0.

EDIT: I see Soroban is replying to this thread as well. He will explain this better if you are still confused.

3. ok. I have worked the question based upon Deadstar's workings and got the following:

4. 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 $x$ m and $3x$ 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: . $A \;=\;7x^2 - 12x + 9$

Find the value of $x$ 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 $a$ and $12-a$.

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

The left piece of wire is bent into a $x\text{-by-}3x\text{ rectangle.}$
Code:
            3x
* - - - - - *
|           |
x |           | x
|           |
* - - - - - *
3x
The area of this rectangle is: . $3x\cdot x \:=\:\boxed{3x^2}$

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

The right piece of wire has length: . $12-8x$
It is bent into a square.
Code:
          s
* - - - *
|       |
s |       | s
|       |
* - - - *
s

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

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

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