Optimization Problem for a Piece of Wire (Please Help!)

[ardentmed]

Hey guys,

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

A piece of 2 m long wire is to be cut into two pieces one of which is to be formed into a circle and the other into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a minimum and (b) a maximum?

So let "L" be the length of a piece of wire; 10-L is the other piece.

So piece one would be:
sin60 = h/(2-L)
So the Area of the triangle is:

A=(1/2)(2-L)(√3/2)(2-L)

So what would P be? Would I just use P=2πr and A=πr^2 ?

I'm at a loss as to how I should finish this question. Would I add the two perimeters and areas and then solve for critical points, then substitute the value back into the expressions to solve for the lengths?

Thanks in advance for all the help guys.

Cheers,
ArdentMed.

 

[HallsofIvy]

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

A piece of 2 m long wire is to be cut into two pieces one of which is to be formed into a circle and the other into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a minimum and (b) a maximum?

So let "L" be the length of a piece of wire; 10-L is the other piece.

Why "10- L"? The problem says the length of the wire is 2m. If one piece is L, the other must be 2- L.

So piece one would be:
sin60 = h/(2-L)
So the Area of the triangle is:

A=(1/2)(2-L)(√3/2)(2-L)

So what would P be? Would I just use P=2πr and A=πr^2 ?

What do you mean by "P"? There has been no mention of "P" so far. Is it the perimeter of the triangle?

An equilateral triangle has all sides the same length. If we call the side length "s" then the perimeter is 3s. If you are making the triangle from the piece of length 2- L (I see you have corrected your "10- L") then 3s= 2- L so that s= (2- L)/3. Yes, it follows that the area is \(\displaystyle \frac{\sqrt{3}}{2}(2- L)^2\)

I'm at a loss as to how I should finish this question. Would I add the two perimeters and areas and then solve for critical points, then substitute the value back into the expressions to solve for the lengths?

You haven't yet said anything about the circle. If you used the wire of length 10- L to form the triangle, then you must use the wire of length L to form the triangle. A circle of radius r has circumference \(\displaystyle 2\pi r\) and area \(\displaystyle \pi r^2\). You are given that \(\displaystyle 2\pi r= L\) so that \(\displaystyle r= \frac{L}{2\pi}\). The area then is \(\displaystyle \pi r^2= \pi \frac{L^2}{4\pi^2}= \frac{L^2}{4\pi}\).

Yes, since you are asked to minimize or maximize the sum of the areas, add the areas (the "sum of the perimeters" is obviously L)"
\(\displaystyle \frac{\sqrt{3}}{2}(2- L)^2+ \frac{L^2}{4\pi}\).

Differentiate that and set it equal to 0, solve for L.

 

[ardentmed]

Thank you so much for the insightful response.

Why "10- L"? The problem says the length of the wire is 2m. If one piece is L, the other must be 2- L.

What do you mean by "P"? There has been no mention of "P" so far. Is it the perimeter of the triangle?

An equilateral triangle has all sides the same length. If we call the side length "s" then the perimeter is 3s. If you are making the triangle from the piece of length 2- L (I see you have corrected your "10- L") then 3s= 2- L so that s= (2- L)/3. Yes, it follows that the area is \(\displaystyle \frac{\sqrt{3}}{2}(2- L)^2\)

You haven't yet said anything about the circle. If you used the wire of length 10- L to form the triangle, then you must use the wire of length L to form the triangle. A circle of radius r has circumference \(\displaystyle 2\pi r\) and area \(\displaystyle \pi r^2\). You are given that \(\displaystyle 2\pi r= L\) so that \(\displaystyle r= \frac{L}{2\pi}\). The area then is \(\displaystyle \pi r^2= \pi \frac{L^2}{4\pi^2}= \frac{L^2}{4\pi}\).

Yes, since you are asked to minimize or maximize the sum of the areas, add the areas (the "sum of the perimeters" is obviously L)"
\(\displaystyle \frac{\sqrt{3}}{2}(2- L)^2+ \frac{L^2}{4\pi}\).

Differentiate that and set it equal to 0, solve for L.

That's strange. I got:

A' = L/(2pi) - √(3)*(2-L)

Which ultimately gave me:

L= 4pi√(3) / 1-2pi√3

How would I go about finding the maximum and minimum for such a function?

Thanks again for the help. I greatly appreciate it.

 

[Ishuda]

...Yes, it follows that the area is \(\displaystyle \frac{\sqrt{3}}{2}(2- L)^2\) ...

Isn't the area \(\displaystyle \frac{\sqrt{3}}{2}s^2\)?

 

[ardentmed]

I'm confused. Either way, how does one obtain the maximum and minimum for this particular problem?

Thanks in advance.

 

[HallsofIvy]

Doing what has already been said, you have a wire of length 2 cut into two pieces, one of length L and the other of length 2- L. The first is bent into an equilateral triangle, with each side of length (2- L)/3. As Ishuda pointed out, correcting me, the area of that triangle would be \(\displaystyle (\sqrt{3}/2)(2- L)^2/9= (\sqrt{3}/18)(2- L)^2\). The other half, of length L, is bent into a circle with circumference \(\displaystyle 2\pi r= L\) so radius \(\displaystyle r= L/(2\pi)\) and area \(\displaystyle \pi r^2= \pi(L^2/4\pi^2)= L^2/4\pi\). The total area is \(\displaystyle A= (\sqrt{3}/18)(2- L)^2+ L^2/4\pi\). To find the value of L that makes that optimum, differentiate with respect to L and set equal to 0.

 

[Ishuda]

I'm confused. Either way, how does one obtain the maximum and minimum for this particular problem?

Thanks in advance.

As HallsofIvy indicated.

First figure out the area function for the triangle as a function of L (it is easier if you actually use (2-L);
A_t(L) = a (L-2)²

Next figure out the area function for the circle
A_c(L) = b L²

The total area is the sum of the triangle and circle areas so add them together
A(L) = a (L-2)² + b L²

Thus the area function is a global maximum (or minimum) where the function derivative is zero or at one of the end points. So take the derivative of the function, set it equal to zero and compute what L needs to be (call that L₁). Evaluate the function at the end points and at L₁. Call those values A₀= A(L₀), A₁=A(L₁), and A₂ = A(L₂). The largest of A₀, A₁, and A₂ is the global maximum, the smallest is the global minimum.

To find out whether A₁ is a local maximum or minimum compute the second derivative of A. If that is positive at L₁ then A₁ is a local minimum, if it is negative then it is a local maximum.

 

[ardentmed]

As HallsofIvy indicated.

First figure out the area function for the triangle as a function of L (it is easier if you actually use (2-L);
A_t(L) = a (L-2)²

Next figure out the area function for the circle
A_c(L) = b L²

The total area is the sum of the triangle and circle areas so add them together
A(L) = a (L-2)² + b L²

Thus the area function is a global maximum (or minimum) where the function derivative is zero or at one of the end points. So take the derivative of the function, set it equal to zero and compute what L needs to be (call that L₁). Evaluate the function at the end points and at L₁. Call those values A₀= A(L₀), A₁=A(L₁), and A₂ = A(L₂). The largest of A₀, A₁, and A₂ is the global maximum, the smallest is the global minimum.

To find out whether A₁ is a local maximum or minimum compute the second derivative of A. If that is positive at L₁ then A₁ is a local minimum, if it is negative then it is a local maximum.

Thanks for the help. But isn't this easier said than done?

I've hit a roadblock at the moment, and am unable to figure out a single value for L let alone two.

I differentiated and multiplied by the common denominator and obtained:

(9L - 4 π√ ̅3 + 2 πL√ ̅3)/ 18 π = 0

Thus,

L = (4 π√ ̅(3) )/ (9+2 πL√ ̅(3))

What is the best course of action to take from here?

Thanks in advance.

 

[Ishuda]

...L = (4 π√ ̅(3) )/ (9+2 π√ ̅(3))...

First of all, sorry I didn't say L₀ was 0 and L₂ was 2. Thus
A₀ = 4a
or, from HallsofIvy's post,
A₀ = (2 / 9 ) √ ̅(3) ~ .4

A₂ = 2b
or
A₂ = 2 / (4 π) ~ .2

The L you determined is L₁ which is about 1 so A₁ is about .2. It looks like the global maximum is A₀ and you will have to use a calculator to actually see what the global min is (but see below). Since both a and b are positive, the second derivative is positive and there is a local minimum at L₁


Note: The approximations given are very rough as I just did the calculations 'off the top of my head'. Also, since the derivative is positive for L greater than L₁, A₂ must be greater than A₁, so the global minimum is at L₁.

 

[ardentmed]

First of all, sorry I didn't say L₀ was 0 and L₂ was 2. Thus
A₀ = 4a
or, from HallsofIvy's post,
A₀ = (2 / 9 ) √ ̅(3) ~ .4

A₂ = 2b
or
A₂ = 2 / (4 π) ~ .2

The L you determined is L₁ which is about 1 so A₁ is about .2. It looks like the global maximum is A₀ and you will have to use a calculator to actually see what the global min is (but see below). Since both a and b are positive, the second derivative is positive and there is a local minimum at L₁


Note: The approximations given are very rough as I just did the calculations 'off the top of my head'. Also, since the derivative is positive for L greater than L₁, A₂ must be greater than A₁, so the global minimum is at L₁.

Thanks again. One final question though, how did you go about finding the other L values? Are those endpoint values?

 

[ardentmed]

Using Area with Respect to Perimeter.

Alright, so knowing that C=2-T, the following substitution can be made:

A = (2-T)^2 / 4[FONT=MathJax_Math-italic]π[/FONT] + (√3*[T^3])/36

Optimizing results in:

A' = (T-2)/2[FONT=MathJax_Math-italic]π[/FONT] + √3(T^2)/12

Thus,

0=T^2 * [FONT=MathJax_Math-italic]π[/FONT]√3 + 6T - 12

Using the quadratic formula, I computed ~1.032742 and ~ -2.135399.

Am I on the right track?

Thanks again.

 

[Ishuda]

Thanks again. One final question though, how did you go about finding the other L values? Are those endpoint values?

Yes, those are endpoints.

 

[Ishuda]

Alright, so knowing that C=2-T, the following substitution can be made:

A = (2-T)^2 / 4[FONT=MathJax_Math-italic]π[/FONT] + (√3*[T^3])/36

Optimizing results in:

A' = (T-2)/2[FONT=MathJax_Math-italic]π[/FONT] + √3(T^2)/12

Thus,

0=T^2 * [FONT=MathJax_Math-italic]π[/FONT]√3 + 6T - 12

Using the quadratic formula, I computed ~1.032742 and ~ -2.135399.

Am I on the right track?

Thanks again.

Yes, you are on the right track assuming your function were correct. We could ignore the answer L~-2.135 because the piece of wire has to have a positive (or at least non-negative) length. However, I believe that 36 in the denominator is supposed to be 18.

 

[ardentmed]

Alright, so I found the endpoints and re-did the question from scratch, taking into account that the equilateral triangle's angle is $\pi$/3 . I also found the endpoints:




I think I'm on the right track. What do you think, Ishuda?


Thanks in advance.

As HallsofIvy indicated.

First figure out the area function for the triangle as a function of L (it is easier if you actually use (2-L);
A_t(L) = a (L-2)²

Next figure out the area function for the circle
A_c(L) = b L²

The total area is the sum of the triangle and circle areas so add them together
A(L) = a (L-2)² + b L²

Thus the area function is a global maximum (or minimum) where the function derivative is zero or at one of the end points. So take the derivative of the function, set it equal to zero and compute what L needs to be (call that L₁). Evaluate the function at the end points and at L₁. Call those values A₀= A(L₀), A₁=A(L₁), and A₂ = A(L₂). The largest of A₀, A₁, and A₂ is the global maximum, the smallest is the global minimum.

To find out whether A₁ is a local maximum or minimum compute the second derivative of A. If that is positive at L₁ then A₁ is a local minimum, if it is negative then it is a local maximum.

 

[ardentmed]

Guys, can you please reply? This question is very urgent.