Another wire problem.

[Flic]

The problem reads as follows.

A wire of length 100cm is cut into 2 pieces. One piece is bent into a circle, the other into a square. Where should the cute be made to maximize the sum of the areas of the square and circle?

This is what I have so far, and I want to know if it is right and what to do next.

(pi)r^2 + x^2 = max (combining the areas)

2(pi)r + 4x = 100 (combining the perimeter)

x = 25-(1/2)(pi)r (solution for X)

(pi)r^2 + (25 - (1/2)(pi)r)^2 (plugging in x to original)

 

[Unco]

G'day, Flic.

Your work is outstanding!

You have the area to maximise being
\(\displaystyle \L A = \pi r^2 + (25 - \frac{1}{2}\pi r)^2\)

When we maximise, we differentiate and set that derivative to zero, right?

It's probably best we expand those parentheses so we can differentiate term by term:

\(\displaystyle \L A = \pi r^2 + 625 - 25\pi r + \frac{1}{4}\pi^2 r^2\)

Now differentiate term by term:

\(\displaystyle \L \pi r^2 \Rightarrow \ 2\pi r\)

\(\displaystyle \L 625 \Rightarrow 0\)

\(\displaystyle \L -25\pi r \Rightarrow -25\pi\)

\(\displaystyle \L \frac{1}{4}\pi^2 r^2 \Rightarrow \frac{1}{2}\pi^2 r\)

And we have

\(\displaystyle \L A' = 2\pi r - 25\pi + \frac{1}{2}\pi^2 r\)

Now set \(\displaystyle A' = 0\):

\(\displaystyle \L 0 = 2\pi r - 25\pi + \frac{1}{2}\pi^2r\)

Do you follow that?

See if you can solve for r, and then determine where the cut was made. The question doesn't ask, but which length goes to which shape?

 

[Flic]

\(\displaystyle \L \frac{1}{4}\pi^2 r^2 \Rightarrow \frac{1}{2}\pi^2 r\)

That is what I was missing, now I shall try to finish it.

\(\displaystyle \L 0 = 2\pi r - 25\pi + \frac{1}{2}\pi^2r\)

\(\displaystyle \L 2r + \frac{1}{2}\pi r = 25\)

\(\displaystyle \L r = \frac{25}{2+\frac{1}{2}\pi}\)

inserting that into circumference formula, I get about 44 (43.99)
solving for x, I got x=14(.0025), inserting that into the 4x, it is 56...

44+56 = 100

44 is the circle, 56 is the square.