Minimum Question

[SeekerOfDragons]

I mainly want to verify my answer to the following question:

A 25-in piece of string is cut into two pieces. One piece is used to form a circle and the other to form a square. How should the string be cut so that the sum of the areas is a minimum? Round to the nearest tenth if necessary.

using the following formulas:
Area of circle = Pi r^2
Circumference = 2 Pi r --> r = Circ / 2 Pi

Area of Square = s^2
Perimeter of Square = 4s

*--------- (X) ---------*--------- (25 - X) -------------*

resulting in a formula of:

Pi * (X/2Pi)^2 + ((25 - X)/4)^2

Finding the derivative of the above:
X/2Pi - (25 - X)/8

setting the above = 0 and solving for x I come up with

X = 50Pi/(8 + 2Pi) which comes to about 10.997521 Inches.

If I did it correctly, I would need a string 11 inches and a string 14 inches.

Did I do this correctly?

 

[galactus]

r=radius of circle and s=side length of square. Let x=length of wire used for circle and y=length of wire used for square.

If we make a circle and a square, then the sum of the enclosed areas is \(\displaystyle A={\pi}r^{2}+s^{2}\) where \(\displaystyle r=\frac{x}{2\pi} \;\ and \;\ s=\frac{y}{4}\)

That's because x is the circumference of the circle and y is the perimeter of the square.

Therefore, \(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{y^{2}}{16}\)

But \(\displaystyle x+y=25, \;\ so \;\ y=25-x\)

\(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{(25-x)^{2}}{16}\)

Set to 0 and solve for x. That will be the minimum.

\(\displaystyle \frac{dA}{dx}=\frac{({\pi}+4)x-25{\pi}}{8{\pi}}\)

\(\displaystyle \frac{({\pi}+4)x-25{\pi}}{8{\pi}}=0\)

\(\displaystyle x=\frac{25{\pi}}{{\pi}+4}\approx 11\)

You were correct and I am sorry for the duh.

 

[SeekerOfDragons]

I'm not really following where you're getting x + y = 12 or y = 12 - x

My string is 25 inches total so in your example X + Y = 25 and y = 25 - x.

 

[BigGlenntheHeavy]

\(\displaystyle P \ = \ 4s+2\pi r \ = \ 25\)

\(\displaystyle A \ = \ s^{2}+\pi r^{2}\)

\(\displaystyle s \ = \ \frac{(25-2\pi r)}{4}\)

\(\displaystyle Hence, \ A \ = \ \bigg[\frac{(25-2\pi r)}{4}\bigg]^{2}+\pi r^{2}\)

\(\displaystyle \frac{dA}{dr} \ = \ \frac{1}{16}(-100\pi+8\pi^{2}r+32\pi r)\)

\(\displaystyle Setting \ the \ slope \ = \ to \ zero, \ one \ gets \ r \ = \ about \ 1.75, \ hence \ s \ = \ about \ 3.5.\)

\(\displaystyle 4(3.5)+2\pi(1.75) \ = \ 25, \ QED\)

\(\displaystyle 4s \ = \ 14 \ and \ 2\pi r \ = \ 11, \ you \ are \ correct \ Seeker \ of \ Dragons.\)

 

[galactus]

I am sorry. I showed how to do a similar problem and it was a 12 inch wire. Just a typo due to transpositional cerebral flatulence.

I changed it accordingly.