Optimization Problems

[Vì sao không tên]

Help me! I have read it many times, but I don't know how to make a relationship between the.

1.
Four feet of wire is to be used to form a square and a circle. How much of wire should be used for the square and how much should be used for the circle in order to enclose the maximum possible area?
2.
Two posts, one 12 f and the other 28f, stand 30 f apart. They are to be stayed by two wires, attached to a single stake, running from around level to the top of each post. Where should the stake be placed to use the least amount of wire ?

Hopefully, see your answer soon!!!

[lehoangel]

In your first question, I think with a limited length of wire, we use to form a circle and with the same wire, we form a square then the area of circle will be more than square. So, we should use all 4 feet of wire to form a circle to have a maximum area.

[Vì sao không tên]

You know, this is not a guessing problem, and the answer always different you imagine. The requirement is you form both circle and square. You have no choice.

[lehoangel]

Yes, the result may be different with my answer. But I don't know your answer so I'll keep my way. Let's see, I can prove my opinion that if we use all amount of wire to form a circle, we'll have a maximum area. Certainly, this area is larger than area of a circle and a square. If the requirement is forming both circle and square, we just use as few as possible amount of wire.. so I think it won't have an exactly answer. How about your answer ?

[Vì sao không tên]

Honestly, I still not get the right answer, or have a connection between them. First u have find the formula, then take derivative to check the max and min. That's why they give you a square and a circle, to make it relate to each other.

However, thanks a lot. You look interested in math!!!

[buidoi_1life1love]

Both of these two problems are pretty simple. Use some calculus to solve it. I'll give you a hint how to solve the first one.
X: the length of the piece of wire that forms the circle.
4-X : the length of the piece of wire that forms the square.
S: the length of a side of the square.
R: radius of the circle.

We have:
X= 2*pi*R ( you should be able to tell why), solve for R, we obtain R=X/ (2*pi)
S= (4-X)/4

Now use the formulas for areas:
Area of the circle= pi*(R^2)= pi*((X/ (2*pi)^2)= (X^2)/(4*pi)
Area of the square= S^2= ((4-X)/4)^2

From here you add the two areas, and see that now the formula for the total area only has one variable, namely X. Taking the derivative of the total area function, then I assume you know how to find max/min. Good luck.

P.S: you should restrict the domain of X and I think you know how to do that.

[Vì sao không tên]

Actually, the second one is the one, I am confused! I can't draw the picture of it. Then I don't know where to start!

Thank you!

[buidoi_1life1love]

You're asking for too much. From what I read above, you and even the other guy don't even have a clue how to approach the first problem. You guys just talked about some random and vague approach, and none is accurate. Now you're telling me that you only have trouble with the second one. If you really needed help with only the second problem, then why did you post two problems and discuss with the other guy how to solve the first one? If you still can't figure out how to the second one after my hint for the first one, just say it. Don't pretend that you knew how to solve the first one already, and then ask for another solution. That's why you never become good at math. You're just lazy to think.

[lehoangel]

I said that I can prove my opinion. And your answer similar mine. Let plus your area of the circle and square then you'll see that it won't have an exactly result.

[buidoi_1life1love]

I said that I can prove my opinion. And your answer similar mine. Let plus your area of the circle and square then you'll see that it won't have an exactly result.
what do you mean by "it won't have have an exactly result."? I'm trying to tell you guys that you guys were not actually solving the problems, and what you said is invalid in math. Show the work in a mathematical way. Don't just make some general statements since math requires accuracy.

[Vì sao không tên]

Hey!! hey!! calm down!! I'm not good in math, but I am not lazy man!!! If you don't feel comfortable to do it, so do not do. I posted those problem to make some interesting topic, also I heed some help.
Like a test, we have some easy and the other harder. However, if no one help me here!!! I will do by myself. I regret that I posted those problems. I should post some other that is really harder.... Actually I don't need to do this stupid thing.


By the way, Thank you for paying attention!!!

[lehoangel]

what do you mean by "it won't have have an exactly result."? I'm trying to tell you guys that you guys were not actually solving the problems, and what you said is invalid in math. Show the work in a mathematical way. Don't just make some general statements since math requires accuracy.

Okie, if you really want me to do that. This is my answer :
--------------------
X: the length of the piece of wire that forms the circle.
4-X : the length of the piece of wire that forms the square.
A: the length of a side of the square.
R: radius of the circle.

X= 2*pi*R
R=X/ (2*pi)
A= (4-X)/4

Area of the circle : C = pi*R^2=pi*(X/2*pi)^2= X^2/(4*pi)
Area of the square : S=A^2=[(4-X)/4]^2=(4-X)^2/16

C+S = X^2/4*pi + (4-X)^(2/16) = [4*X^2 + pi*(4-X)^2]/(16*pi) = [(4+pi)*X^2-8*pi*X+16*pi]/(16*pi)

If we call f(x)= [(4+pi)*X^2-8*pi*X+16*pi]/(16*pi)
then f(x)' > 0 (because X>0) => f(x) is an increase function.
So, when value of X is increasing to 4, value of f(x) is also increasing to 4/pi.

It means if we use all length of wire to form a circle, we'll have a maximum area !

[Vì sao không tên]

My eyes are been burning by watching your solution. "_" I have my answer now. don't make my topic become serious guys.

[lehoangel]

Oh nice :) Can you show me your answer ?

[Vì sao không tên]

OK, I am not sure it is right ! we have 2.24 ... for circle, then check again..... the max Area, It should be 4