# problem with finding the minimum area

• Jun 6th 2009, 08:05 PM
helloying
problem with finding the minimum area
A piece of wire of length 8cm is cut into two pieces,one of length x cm,the other of length (8-x)cm.The piece of lenght x cm is bent to form a circle with circumference x cm. The other piece is bent to form a square with perimeter (8-x)cm. Show that , as x varies, the sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the circle is$\displaystyle \frac{4}{4+\pi}$

i found the area of the square to be $\displaystyle (4-\frac{x}{2})cm^2$
and the the radius of the circle is $\displaystyle \frac{x}{2\pi}$so the area of circle is $\displaystyle \frac{x^2}{4\pi}$

i add them up and i differentiate the eqn. i got $\displaystyle \frac{x}{2\pi}-0.5$

At min point,$\displaystyle \frac{x}{2\pi}-0.5$=0
thus x=$\displaystyle \frac{1}{\pi}$

then when i sub it into the radius, i did not get the ans. what is wrong?
• Jun 6th 2009, 10:12 PM
Prove It
Quote:

Originally Posted by helloying
A piece of wire of length 8cm is cut into two pieces,one of length x cm,the other of length (8-x)cm.The piece of lenght x cm is bent to form a circle with circumference x cm. The other piece is bent to form a square with perimeter (8-x)cm. Show that , as x varies, the sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the circle is$\displaystyle \frac{4}{4+\pi}$

i found the area of the square to be $\displaystyle (4-\frac{x}{2})cm^2$
and the the radius of the circle is $\displaystyle \frac{x}{2\pi}$so the area of circle is $\displaystyle \frac{x^2}{4\pi}$

i add them up and i differentiate the eqn. i got $\displaystyle \frac{x}{2\pi}-0.5$

At min point,$\displaystyle \frac{x}{2\pi}-0.5$=0
thus x=$\displaystyle \frac{1}{\pi}$

then when i sub it into the radius, i did not get the ans. what is wrong?

The circumference of the circle is $\displaystyle x\,\textrm{cm}$, so

$\displaystyle x = 2\pi r$

$\displaystyle r = \frac{x}{2\pi}\,\textrm{cm}$.

The area of the circle is $\displaystyle \pi r^2 = \pi \left( \frac{x}{2\pi} \right) ^2 = \frac{x^2}{4\pi}\,\textrm{cm}^2$.

The Perimeter of the square is $\displaystyle (8 - x)\,\textrm{cm}$.

So $\displaystyle 8 - x = 4l$

$\displaystyle l = \left(2 - \frac{1}{4}x\right)\,\textrm{cm}$.

Therefore the Area of the Square is $\displaystyle \left(2 - \frac{1}{4}x\right)^2 = 4 - x + \frac{1}{16}x^2$.

So the Area of both shapes together is given by...

$\displaystyle A = \frac{1}{4\pi}x^2 + 4 - x + \frac{1}{16}x^2$

$\displaystyle = \left(\frac{1}{4\pi} + \frac{1}{16}\right)x^2 - x + 4$

$\displaystyle = \frac{4 + \pi}{16\pi}x^2 - x + 4$.

To find the minimum area, we take the derivative, set it equal to 0, and solve for x.

$\displaystyle \frac{dA}{dx} = \frac{4 + \pi}{8\pi}x - 1$

$\displaystyle 0 = \frac{4 + \pi}{8\pi}x - 1$

$\displaystyle 1 = \frac{4 + \pi}{8\pi}x$

$\displaystyle x = \frac{8\pi}{4 + \pi}$.

We can show we have a minimum by showing that it's second derivative is positive at that point...

$\displaystyle \frac{d^2A}{dx^2} = \frac{4 + \pi}{8\pi} > 0$.

Since the second derivative is positive at that point (and all points), we have a minimum.

So the minimum occurs when $\displaystyle x = \frac{8\pi}{4 + \pi}\,\textrm{cm}$.

Remember that $\displaystyle r = \frac{x}{2\pi}\,\textrm{cm}$

$\displaystyle r = \frac{\frac{8\pi}{4 + \pi}}{2\pi}$

$\displaystyle = \frac{8\pi}{4 + \pi} \times \frac{1}{2\pi}$

$\displaystyle = \frac{4}{4 + \pi}\,\textrm{cm}$.