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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?
 2 years ago
 2 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?
 2 years ago
 2 years ago

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satellite73Best ResponseYou've already chosen the best response.0
if you want a guess i would say put everything in to the circle and forget about the square
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
oh sorry you already answered that
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
probably all in the square to minimize rigth? just a guess though
 2 years ago

ktklownBest ResponseYou've already chosen the best response.1
that's my guess too, but, i'll bring mathematica out again ;)
 2 years ago

ktklownBest ResponseYou've already chosen the best response.1
yeah, mathematica says to put 100% into the square
 2 years ago

ktklownBest ResponseYou've already chosen the best response.1
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)}}
 2 years ago

hersheys06Best ResponseYou've already chosen the best response.0
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
 2 years ago

hersheys06Best ResponseYou've already chosen the best response.0
it says it's incorrect :S
 2 years ago

ktklownBest ResponseYou've already chosen the best response.1
oh now i get x=3.5, that's one side of the square
 2 years ago

hersheys06Best ResponseYou've already chosen the best response.0
sorrry, still incorrect T___T;
 2 years ago

ktklownBest ResponseYou've already chosen the best response.1
well it's 4 times that i just gave you one side
 2 years ago

hersheys06Best ResponseYou've already chosen the best response.0
ohh, i did the area (3.5)^2 instead of (3.5)*4. thanksss
 2 years ago

ktklownBest ResponseYou've already chosen the best response.1
did it say that was right?
 2 years ago

robtobeyBest ResponseYou've already chosen the best response.1
(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.
 2 years ago

robtobeyBest ResponseYou've already chosen the best response.1
Take 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.
 2 years ago

robtobeyBest ResponseYou've already chosen the best response.1
Using Mathematica's Minimize function:\[\text{Minimize}\left[ \left\{\left(\frac{x}{4}\right)^2+\frac{(25x)^2}{4 \pi }\right\},x\right]\to \]\[\left\{\frac{2500+\frac{40000}{(4+\pi )^2}+\frac{10000 \pi }{(4+\pi )^2}\frac{20000}{4+\pi }}{16 \pi },\left\{x\to \frac{100}{4+\pi }\right\}\right\}\text{ //N} \]
 2 years ago
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