# Problems with word problems

#### MrShrip

I've come across what I think is an interesting word problem: "With a hundred inches of ink I draw a square and a circle so that he sum of the areas is a minimum. What is the length of the perimeter of the square?".

Now, I feel like I more or less have the set-up down, e.g. take the sum of the two perimeters and solve for the radius of the circle, substitute this new found value of r into the sum of the areas followed by a load of algebra and a few derivatives, yadda yadda, ect ect (Math for this part below.) I played around with these numbers for a while until I came to what I thought was the solution. It wasn't. I ran the math a few more times and with each attempt I came to a different result. At this point I'm not sure if the set-up is wrong or if I'm making a mistake somewhere in the minutiae of the algebra. A little bit of help would be much appreciated.

4x+2πr = 100 (x equals one side of the square.)r = 100∕(4x+2π)

A = x²+πr² ("A" equals the sum of the areas of the square and circle.)
A=x²+π(100/(4x+2π))².....

#### romsek

MHF Helper
$s \text{ is the length of a side of the square.}$

$r \text{ is the radius of the circle.}$

The sum of the perimeters is 100 inches.
$4s + 2\pi r = 100$

The area of the two shapes
$A = \pi r^2 + s^2$

Solving the constraint equation isn't too difficult so we won't need to use Lagrange multipliers.

$r = \dfrac {2}{\pi} (25-s)$

$A = \pi \left(\dfrac {2}{\pi} (25-s)\right)^2 + s^2=\dfrac 1 \pi((\pi+4)s^2 -200s + 2500$

Now we find the critical point of $A$ as usual

$\dfrac{dA}{ds} =\left(2+\dfrac{8}{\pi }\right) s-\dfrac{200}{\pi }$

$\dfrac{dA}{ds} =0 \Rightarrow s=\dfrac{100}{4+\pi}$

$\dfrac{d^2A}{{ds}^2} = 2+\dfrac 8 \pi > 0$ so this represents a minimum of $A$

The length of the perimeter of the square is just

$P_{square}=4s = \dfrac{400}{4+\pi}\approx 56.01$

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2 people

#### skeeter

MHF Helper
I've come across what I think is an interesting word problem: "With a hundred inches of ink I draw a square and a circle so that he sum of the areas is a minimum. What is the length of the perimeter of the square?".
$x$ = square perimeter

$100-x$ =circle circumference

$A = \dfrac{x^2}{16} + \dfrac{(100-x)^2}{4\pi}$

$\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{100-x}{2\pi}$

set $\dfrac{dA}{dx} = 0$ ...

$\dfrac{x}{8} = \dfrac{100-x}{2\pi}$

$x = \dfrac{400}{\pi+4} \approx 56$ inches

$\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{1}{2\pi} > 0$ for all $0 \le x \le 100$

so $x = \dfrac{400}{\pi+4}$ is the perimeter of the square with minimum total area

2 people