Optimizing Area for two shapes using a single piece of string

[ardentmed]

Hey guys,

I would really appreciate some help for this question I'm stuck on at the moment:

"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?"

The the area of that triangle would be (3√/2)(2−L)2/9=(3√/18)(2−L)2. The other half, of length L, is bent into a circle with circumference 2πr=L so radius r=L/(2π) and area πr2=π(L2/4π2)=L2/4π. The total area is A=(3√/18)(2−L)2+L2/4π

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.

 

[MarkFL]

You may find this post helpful:

http://mathhelpboards.com/questions-other-sites-52/s-questions-yahoo-answers-regarding-optimization-5221.html#post23754

 

[ardentmed]

So I would have to use Lagrange Multipliers in order to find the other extrema?

Why is the interval 0<C<L?

Thanks in advance.

 

[MarkFL]

No, you should wind up with a quadratic function, and so there are 3 ways I gave in that other thread to carry out the optimization...a pre-calculus method, a single-variable calculus method, or a multi-variable method (Lagrange multipliers). Given the fact that you are not studying Calc III yet, I would focus on the single-variable calculus method.

Well, the amount of string used to make up either shape must range from zero up to the length of the string, right?

 

[MarkFL]

Let's ignore that other thread for now, and work this problem in general terms.

Let the length of string be $L$, the portion formed into a circle be $C$ and the portion formed into an equilateral triangle be $T$. So, we must have:

\(\displaystyle C+T=L\)

Now, we need the area of a circle as a function of its circumference, and we need the area of an equilateral triangle as a function of its perimeter. Can you state these?

 

[ardentmed]

Let's ignore that other thread for now, and work this problem in general terms.

Let the length of string be $L$, the portion formed into a circle be $C$ and the portion formed into an equilateral triangle be $T$. So, we must have:

\(\displaystyle C+T=L\)

Now, we need the area of a circle as a function of its circumference, and we need the area of an equilateral triangle as a function of its perimeter. Can you state these?

If Pc = 2$\pi$r, then r= P/2$\pi$

Thus, Ac = $\pi$(P/2$\pi$)^2

And if Pt = 3L

Then At = P/3 * 1/2 * √ (3)/2

Am I on the right track?

Thanks again.

 

[MarkFL]

If we let $A_C$ be the area form by the circle and $A_T$ be the area formed by the triangle, then we may proceed as follows:

\(\displaystyle A_C=\pi r^2=\pi\left(\frac{C}{2\pi}\right)^2=\frac{C^2}{4\pi}\)

\(\displaystyle A_T=\frac{1}{2}s^2\sin\left(60^{\circ}\right)=\frac{1}{2}\cdot\left(\frac{T}{3}\right)^2\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}T^2}{36}\)

And so the total area $A$ formed by the two shapes is:

\(\displaystyle A=A_C+A_T=\frac{C^2}{4\pi}+\frac{\sqrt{3}T^2}{36}\)

Now, how can we express $A$ as a function of one variable?

 

[ardentmed]

If we let $A_C$ be the area form by the circle and $A_T$ be the area formed by the triangle, then we may proceed as follows:

\(\displaystyle A_C=\pi r^2=\pi\left(\frac{C}{2\pi}\right)^2=\frac{C^2}{4\pi}\)

\(\displaystyle A_T=\frac{1}{2}s^2\sin\left(60^{\circ}\right)=\frac{1}{2}\cdot\left(\frac{T}{3}\right)^2\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}T^2}{36}\)

And so the total area $A$ formed by the two shapes is:

\(\displaystyle A=A_C+A_T=\frac{C^2}{4\pi}+\frac{\sqrt{3}T^2}{36}\)

Now, how can we express $A$ as a function of one variable?

Knowing that the total perimeter of the two shapes is equal to 10, we can add the perimeters together and use the perimeter function as a system of equations.

Thus, we can isolate either variable on one side, and use the resulting value and insert it into the area function and then proceed to take the derivative thereof to solve for the remaining critical points.

 

[MarkFL]

I would stick to the general length, and use:

\(\displaystyle C+T=L\)

and substitute to remove one of the two variables, and then optimize. (Yes)

 

[ardentmed]

I would stick to the general length, and use:

\(\displaystyle C+T=L\)

and substitute to remove one of the two variables, and then optimize. (Yes)

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

A = (2-T)^2 / 4$\pi$ + (√3*[T^3])/36

Optimizing results in:

A' = (T-2)/2$\pi$ + √3(T^2)/12

Thus,

0=T^2 * $\pi$√3 + 6T - 12

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

Am I on the right track?

Thanks again.

 

[MarkFL]

How did you get a cubic in $T$?

 

[ardentmed]

How did you get a cubic in $T$?

Thanks for catching my mistake. I made a slight typo at the beginning of my work. I ended up getting:

0= (18T-36+√3*T)/36π

T=36/(18+√3)

T~1.824443.

Am I on the right track?

Thanks again.

 

[MarkFL]

Okay, this is how I would proceed from where I left off:

\(\displaystyle A=\frac{C^2}{4\pi}+\frac{\sqrt{3}T^2}{36}\)

Using $T=L-C$, we have:

\(\displaystyle A(C)=\frac{C^2}{4\pi}+\frac{\sqrt{3}(L-C)^2}{36}\)

Differentiating, and equating the result to zero, we find:

\(\displaystyle A'(C)=\frac{C}{2\pi}-\frac{\sqrt{3}}{18}(L-C)=\frac{2\sqrt{3}\left(\left(\pi+3\sqrt{3}\right)C-\pi L\right)}{36\pi}=0\)

This implies:

\(\displaystyle \left(\pi+3\sqrt{3}\right)C-\pi L=0\)

Solving for $C$, we obtain:

\(\displaystyle C=\frac{\pi L}{\pi+3\sqrt{3}}\)

Observing that $A''(C)$ is a positive constant, we know this critical value is at the global minimum. And so we find:

\(\displaystyle T=L-\frac{\pi L}{\pi+3\sqrt{3}}=\frac{3\sqrt{3}L}{\pi+3\sqrt{3}}\)

Using $L=2$ gives a different result than what you posted. Can you show your work?

 

[ardentmed]

Okay, this is how I would proceed from where I left off:

\(\displaystyle A=\frac{C^2}{4\pi}+\frac{\sqrt{3}T^2}{36}\)

Using $T=L-C$, we have:

\(\displaystyle A(C)=\frac{C^2}{4\pi}+\frac{\sqrt{3}(L-C)^2}{36}\)

Differentiating, and equating the result to zero, we find:

\(\displaystyle A'(C)=\frac{C}{2\pi}-\frac{\sqrt{3}}{18}(L-C)=\frac{2\sqrt{3}\left(\left(\pi+3\sqrt{3}\right)C-\pi L\right)}{36\pi}=0\)

This implies:

\(\displaystyle \left(\pi+3\sqrt{3}\right)C-\pi L=0\)

Solving for $C$, we obtain:

\(\displaystyle C=\frac{\pi L}{\pi+3\sqrt{3}}\)

Observing that $A''(C)$ is a positive constant, we know this critical value is at the global maximum. And so we find:

\(\displaystyle T=L-\frac{\pi L}{\pi+3\sqrt{3}}=\frac{3\sqrt{3}L}{\pi+3\sqrt{3}}\)

Using $L=2$ gives a different result than what you posted. Can you show your work?

Alright, so if T ~ 1.24, that gives us one of the areas, then,

This represents the maximum total area, since this was the global maximum. As this function has no absolute maximum (or minimum?), the maximum (and minimum?) total area must occur at an endpoint (either L = 0 or L = 2). Checking, we find that L=2 provides the optimal area.

So should I use L=0 for the minimum area (which does not seem viable, but it is also an endpoint), or did I already find that when computing L1 = 1.0947 (which may have been a global minimum if I recall correctly).

 

[MarkFL]

I made a typo above, which I have fixed...a positive second derivative implies a minimum.

So, you then need to look at the boundaries, which occur when either $C$ or $T$ are zero. Now given that a circle will enclose more area than any other closed curve in 2D, where do you expect the maximum to be? (You should check to be sure, but it is nice when intuition agrees with result.)

 

[ardentmed]

Edited for brevity.

 

[ardentmed]

I made a typo above, which I have fixed...a positive second derivative implies a minimum.

So, you then need to look at the boundaries, which occur when either $C$ or $T$ are zero. Now given that a circle will enclose more area than any other closed curve in 2D, where do you expect the maximum to be? (You should check to be sure, but it is nice when intuition agrees with result.)

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, Mark?

Thanks in advance.

 

[ardentmed]

So just for the record, in a nutshell, I optimized it, figured out that it can't be minimum AND maximum at the same point, so I tested the endpoints as well (since, after all, it is a continuous function) at x=0 and x=2 (and x=~.7 for the optimized value).