Question

A piece of wire 25 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.

(a) How much of the wire should go to the square to maximize the total area enclosed by both figures?

ANS= 0 m

*** (b) How much of the wire should go to the square to minimize the total area enclosed by both figures?

Answer 1

if you want a guess i would say put everything in to the circle and forget about the square

Answer 2

oh sorry you already answered that

Answer 3

probably all in the square to minimize rigth? just a guess though

Answer 4

that's my guess too, but, i'll bring mathematica out again ;-)

Answer 5

yeah, mathematica says to put 100% into the square

Answer 6

In[940]:= Minimize[{\[Pi]r^2 + x^2, 4 x + 2 \[Pi]r == 25}, {r, x}]

Out[940]= {1/16 (625 - 100 \[Pi]r + 20 \[Pi]r^2), {r -> 0,
x -> 1/4 (25 - 2 \[Pi]r)}}

Answer 7

How much of the wire should go to the square to minimize the total area enclosed by both figures?

i think it requires a numerical answer

Answer 8

all of it. 25 m

Answer 9

it says it's incorrect :S

Answer 10

hmm

Answer 11

oh now i get x=3.5, that's one side of the square

Answer 12

sorrry, still incorrect T___T;

Answer 13

well it's 4 times that
i just gave you one side

Answer 14

ohh, i did the area (3.5)^2 instead of (3.5)*4. thanksss

Answer 15

did it say that was right?

Answer 16

yess

Answer 17

(b) How much of the wire should go to the square to minimize the total area enclosed by both figures?

The exact answer is:\[\frac{100}{4+\pi } \text{ meters}\]or 14.002479 meters to 8 digits.

Answer 18

Take the derivative of the expression of interest\[D\left[\left(\frac{x}{4}\right)^2+\frac{(25-x)^2}{4 \pi },x\right] \]set it to zero\[-\frac{25-x}{2 \pi }+\frac{x}{8}==0 \]and solve for x.\[x\to \frac{100}{4+\pi } \]A plot is attached.

Answer 19

Using Mathematica's Minimize function:\[\text{Minimize}\left[ \left\{\left(\frac{x}{4}\right)^2+\frac{(25-x)^2}{4 \pi }\right\},x\right]\to \]\[\left\{\frac{2500+\frac{40000}{(4+\pi )^2}+\frac{10000 \pi }{(4+\pi )^2}-\frac{20000}{4+\pi }}{16 \pi },\left\{x\to \frac{100}{4+\pi }\right\}\right\}\text{ //N} \]