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anonymous
 5 years ago
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?
anonymous
 5 years ago
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
 5 years ago
Best ResponseYou've already chosen the best response.0if you want a guess i would say put everything in to the circle and forget about the square

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh sorry you already answered that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0probably all in the square to minimize rigth? just a guess though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's my guess too, but, i'll bring mathematica out again ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, mathematica says to put 100% into the square

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In[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
 5 years ago
Best ResponseYou've already chosen the best response.0How 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
 5 years ago
Best ResponseYou've already chosen the best response.0it says it's incorrect :S

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh now i get x=3.5, that's one side of the square

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorrry, still incorrect T___T;

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well it's 4 times that i just gave you one side

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh, i did the area (3.5)^2 instead of (3.5)*4. thanksss

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did it say that was right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(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
 5 years ago
Best ResponseYou've already chosen the best response.0Take the derivative of the expression of interest\[D\left[\left(\frac{x}{4}\right)^2+\frac{(25x)^2}{4 \pi },x\right] \]set it to zero\[\frac{25x}{2 \pi }+\frac{x}{8}==0 \]and solve for x.\[x\to \frac{100}{4+\pi } \]A plot is attached.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Using Mathematica's Minimize function:\[\text{Minimize}\left[ \left\{\left(\frac{x}{4}\right)^2+\frac{(25x)^2}{4 \pi }\right\},x\right]\to \]\[\left\{\frac{2500+\frac{40000}{(4+\pi )^2}+\frac{10000 \pi }{(4+\pi )^2}\frac{20000}{4+\pi }}{16 \pi },\left\{x\to \frac{100}{4+\pi }\right\}\right\}\text{ //N} \]
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