# Math Help - Differentiation

1. ## Differentiation

Q:

A piece of wire of length d units is cut into 2 pieces. One piece is bent into the form of a square and the other piece is bent into the form of a circle. Prove that, when the combined area of the 2 shapes is a minimum, then the radius of the circle is approximately 0.07d units.

Thank you for helping!

2. Hello

Originally Posted by Tangera
Q:

A piece of wire of length d units is cut into 2 pieces. One piece is bent into the form of a square and the other piece is bent into the form of a circle. Prove that, when the combined area of the 2 shapes is a minimum, then the radius of the circle is approximately 0.07d units.

Thank you for helping!
See here http://www.mathhelpforum.com/math-he...n-problem.html for a similar problem.

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Let m be the length of the wire that will be transformed into a square.
Then, d-m will be the length of the wire that will be transformed into a circle.

The perimeter of the square is m. Therefore, one side of the square is $\frac m4$ long.

The area is known to be the square of the side, that is to say :

$\mathcal{A}_s=\left(\frac m4\right)^2=\frac{m^2}{16}$

The perimeter of the circle is d-m. We know that the perimeter of a circle is $2 \pi R$.

--> $d-m=2 \pi R \implies R=\frac{d-m}{2 \pi}$

The area of a circle is $\mathcal{A}_c=\pi R^2=\pi \left(\frac{d-m}{2 \pi}\right)^2=\frac{(d-m)^2}{4 \pi}$

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2 areas are combined :

$\mathcal{A}(m)=\mathcal{A}_s+\mathcal{A}_c$

m is the variable

3. Originally Posted by Tangera
Q:

A piece of wire of length d units is cut into 2 pieces. One piece is bent into the form of a square and the other piece is bent into the form of a circle. Prove that, when the combined area of the 2 shapes is a minimum, then the radius of the circle is approximately 0.07d units.

Thank you for helping!
It isn't necessary to help you anymore - Moo did all the interesting parts.

So only for confirmation: $R=\frac1{2\pi+8}\cdot d \approx 0.07001239... d$