Minimum area of circle, square from 49" string cut in two

[AGlas9837]

A 49-inch 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?

Is this the correct formula for solving?

S = (49-x/4)^2 + pi(x/2pi)^2

 

[royhaas]

Re: Minimum area

The string is cut at length \(\displaystyle x\) from one end. Then the circumference of the circle is \(\displaystyle \pi x\), while the side of the square is \(\displaystyle \frac {49-x} {4}\).

 

[galactus]

Your set up looks correct. Good work.


..... \(\displaystyle \;\ \;\ \text{x} \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \text{49-x}\)

____________________*_____________________

.......\(\displaystyle \uparrow{\bigcirc}\).......................\(\displaystyle \uparrow{\square}\)


The side of each square will be \(\displaystyle (\frac{49-x}{4})\), so the area is \(\displaystyle (\frac{49-x}{4})^{2}\)

The circumference of the circle is x, so the area is \(\displaystyle \frac{x^{2}}{4\pi}\)

Total area: \(\displaystyle A=(\frac{49-x}{4})^{2}+\frac{x^{2}}{4\pi}\)

Differentiate: \(\displaystyle A'(x)=\frac{({\pi}+4)x-49{\pi}}{8\pi}\)

Set to 0 and solve for x and we see \(\displaystyle x=\frac{49\pi}{\pi+4}\approx{21.555}\)

Plugging back into the area formulas we get \(\displaystyle \text{area of square=}\frac{2401}{(\pi+4)^{2}}\approx{47.076}\)

\(\displaystyle \text{area of circle=}\frac{2401\pi}{4(\pi+4)^{2}}\approx{36.974}\)

 

[AGlas9837]

Thanks! The only problem is I'm not seeing how I get the length of string for each. Multiplying x times pi gives a number that is greater than 60.

 

[Deleted member 4993]

Thanks! The only problem is I'm not seeing how I get the length of string for each. Multiplying x times pi gives a number that is greater than 60.

Why would you multiply 'x' by 'pi'?

I think, you are not writing this problem down - just staring at it!!

 

[Dr. Flim-Flam]

let x = circumference of the circle and 49-x = perimeter of square.

A =Pi(r)^2 + s^2, r = radius of circle and s = side of square.

C = 2*Pi*r, x = 2*Pi*r, r =x/2(Pi) Perimeter = 4s, 49-x = 4s, s = (49-x)/4

A(x) =PI(x^2/4(Pi)^2 + [(49-x)^2]/16 = x^2/4(Pi) +[49-x)^2]/16

A ' (x) = 2x/(4Pi) + [2(49-x)(-1)] /16 = x/2Pi + (x-49)/8.

Setting slope = to zero: x(4+Pi) = 49Pi, x = 49Pi/(4+Pi) Check: x+ 49-x = 21.555+ 27.444 = 48.999999 = 49

Hence A(x) = A[49P1/(4+Pi)] = 84.0498 = min area

 

[AGlas9837]

Yes, apparently I was staring at it in a sleep-deprived stupor. I figured it out with no problem the next day. Sorry, and thanks for everyone's help!