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?

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- anonymous

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

- anonymous

oh sorry you already answered that

- anonymous

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

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## More answers

- anonymous

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

- anonymous

yeah, mathematica says to put 100% into the square

- anonymous

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)}}

- anonymous

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

- anonymous

all of it. 25 m

- anonymous

it says it's incorrect :S

- anonymous

hmm

- anonymous

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

- anonymous

sorrry, still incorrect T___T;

- anonymous

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

- anonymous

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

- anonymous

did it say that was right?

- anonymous

yess

- anonymous

(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.

- anonymous

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.

##### 1 Attachment

- anonymous

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

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