A community for students.
Here's the question you clicked on:
 0 viewing
hersheys06
 3 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?
hersheys06
 3 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?

This Question is Closed

satellite73
 3 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry you already answered that

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0probably all in the square to minimize rigth? just a guess though

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

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

ktklown
 3 years ago
Best ResponseYou've already chosen the best response.1In[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)}}

hersheys06
 3 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

hersheys06
 3 years ago
Best ResponseYou've already chosen the best response.0it says it's incorrect :S

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

hersheys06
 3 years ago
Best ResponseYou've already chosen the best response.0sorrry, still incorrect T___T;

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

hersheys06
 3 years ago
Best ResponseYou've already chosen the best response.0ohh, i did the area (3.5)^2 instead of (3.5)*4. thanksss

ktklown
 3 years ago
Best ResponseYou've already chosen the best response.1did it say that was right?

robtobey
 3 years ago
Best 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.

robtobey
 3 years ago
Best ResponseYou've already chosen the best response.1Take 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.

robtobey
 3 years ago
Best ResponseYou've already chosen the best response.1Using Mathematica's Minimize function:\[\text{Minimize}\left[ \left\{\left(\frac{x}{4}\right)^2+\frac{(25x)^2}{4 \pi }\right\},x\right]\to \]\[\left\{\frac{2500+\frac{40000}{(4+\pi )^2}+\frac{10000 \pi }{(4+\pi )^2}\frac{20000}{4+\pi }}{16 \pi },\left\{x\to \frac{100}{4+\pi }\right\}\right\}\text{ //N} \]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.