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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?

Mathematics
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if you want a guess i would say put everything in to the circle and forget about the square
oh sorry you already answered that
probably all in the square to minimize rigth? just a guess though

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Other answers:

that's my guess too, but, i'll bring mathematica out again ;-)
yeah, mathematica says to put 100% into the square
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)}}
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
all of it. 25 m
it says it's incorrect :S
hmm
oh now i get x=3.5, that's one side of the square
sorrry, still incorrect T___T;
well it's 4 times that i just gave you one side
ohh, i did the area (3.5)^2 instead of (3.5)*4. thanksss
did it say that was right?
yess
(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.
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
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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