I'm stuck! Please help, Pre-Cal Problem

[tabitha.halford]

Hi everyone! Here is the problem that I am trying to solve:

You have a wire that is 26 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

Here is what I have done: I know that the wire will be cut into two sections, one called "x" and the other called "26-x". I know that x will equal circumference, so I set x=2(pi)(r) and solved for r, and plugged this into the area of a circle. I have tried for the last 2 days to solve this problem, I have used derivatives, systems of equations, etc. but this seems to be the closest I have gotten yet. Please help! I would greatly appreciate it!

 

[Deleted member 4993]

Hi everyone! Here is the problem that I am trying to solve:

You have a wire that is 26 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

Here is what I have done: I know that the wire will be cut into two sections, one called "x" and the other called "26-x". I know that x will equal circumference, so I set x=2(pi)(r) and solved for r, and plugged this into the area of a circle. I have tried for the last 2 days to solve this problem, I have used derivatives, systems of equations, etc. but this seems to be the closest I have gotten yet. Please help! I would greatly appreciate it!

Then each side of the square is = (26-x)/4

Area of the square is = (26-x)²/16

area of the circle is = π/4 * (2r)²

Then

A = π/4 * x²/π² + (26-x)²/16

Now minimize A and solve for 'x'.

 

[tabitha.halford]

Thanks, but I am still not getting it right!

I plugged everything in and I ended up with an imaginary number. It was x=-(-3.25)+or-(sqrt((-3.25)²-4(.1421)(42.25))/2(.1421). Is there any other way to solve it?

 

[HallsofIvy]

I plugged everything in and I ended up with an imaginary number. It was x=-(-3.25)+or-(sqrt((-3.25)²-4(.1421)(42.25))/2(.1421). Is there any other way to solve it?

It's impossible to tell what you did wrong if you don't show us what you did! Doing what Subhotosh Kahn suggested, differentiating \(\displaystyle A= \frac{x^2}{4\pi}- \frac{(26- x)^2}{16}\), and setting that equal to 0, gives you a linear equation to solve for x. That cannot possibly have a complex (the value you give is not imaginary) answer.

 

[Deleted member 4993]

I plugged everything in and I ended up with an imaginary number. It was x=-(-3.25)+or-(sqrt((-3.25)²-4(.1421)(42.25))/2(.1421). Is there any other way to solve it?

Do you know how to find derivatives of a function?

If don't, then,

A = π/4 * x²/π² + (26-x)²/16

is a parabola like

y = M*(x-h)² + k

or

y = P*x² + Q*x + R

Where is the vertex of this parabola?