Multivariable Optimization: "A piece of wire 8m long is cut into two parts..."

[mh690]

Multivariable Optimization: "A piece of wire 8m long is cut into two parts..."

Here is the problem I'm struggling with:

A piece of wire 8m long is cut into two parts one of which is bent into a square and the other bent into a circle. Find the radius of the circle if the sum of their areas is to be a minimum.

The equation I made is Area= x^2+pi*r^2
(x being the length/width of the square and r being circle radius)
I used partial differentiation to achieve the first derivatives of each variable which came out the be:
dA/dx= 2x + pi
dA/dr = 2*pi*r

But now I'm unsure what to do with these equations and how I'm supposed to achieve numerical values.
I was given the answer to be r= 0.56 but no steps to solve it.

Thank you!

 

[HallsofIvy]

Here is the problem I'm struggling with:

A piece of wire 8m long is cut into two parts one of which is bent into a square and the other bent into a circle. Find the radius of the circle if the sum of their areas is to be a minimum.

The equation I made is Area= x^2+pi*r^2
(x being the length/width of the square and r being circle radius)
I used partial differentiation to achieve the first derivatives of each variable which came out the be:
dA/dx= 2x + pi
dA/dr = 2*pi*r

But now I'm unsure what to do with these equations and how I'm supposed to achieve numerical values.
I was given the answer to be r= 0.56 but no steps to solve it.

Thank you!

Surely it occurred to you that you should use the fact that "a piece of wire 8m long is cut into two parts"?
You wil need to use the formulas for circumference of a circle and perimeter of a square:
the circumference of a circle of radius r is \(\displaystyle 2\pi r\) and the perimeter of a square with side length x is 4x.
The fact that the two are made from "a piece of wire 8 m long" means that \(\displaystyle 2\pi r+ 4x= 8\).

One thing you can do that is solve that last equation for x: \(\displaystyle x= 2- \pi r/4\).
So \(\displaystyle x^2+ \pi r^2= (2- \pi r/4)^2+ \pi r^2= 4- \pi r+ \pi^2 r^2/16+ \pi r^2\). Set the derivative of that, with respect to r, equal to 0.