Maximum-minimum area from a fixed length rope

[Karol]

Homework Statement

Homework Equations

Area of triangle=(base x height)/2

The Attempt at a Solution

a is the side of the triangle.
Area of an equilateral triangle: $~\displaystyle \frac{a}{2h}=\tan 30^0=\frac{\sqrt{3}}{3}$
$$\rightarrow A_t=\frac{\sqrt{3}}{4}a^2$$
Side of the rectangle: (L-3a)
Total area:
$$A=(L-3a)^2+\frac{\sqrt{3}}{4}a^2$$
$$A'=2(L-3a)(-3)+\frac{\sqrt{3}}{2}a$$
$$A'=0:~a=\frac{12L}{36+\sqrt{3}}$$
There is no other value for the A'=0

 

[kuruman]

What happens if instead of forming the triangle first, you form the square first and $a$ is the side of the square?

 

[LCKurtz]

Remember, max and min points can happen not just at critical points inside the interval, but at end points too.

 

[Karol]

$$A=a^2+\frac{\sqrt{3}}{4}(L-4a)^2$$
$$a=\frac{\sqrt{3}L}{4\sqrt{3}+1}$$

 

[Ray Vickson]

$$A=a^2+\frac{\sqrt{3}}{4}(L-4a)^2$$
$$a=\frac{\sqrt{3}L}{4\sqrt{3}+1}$$

This is not a useful answer, since it does not specify whether your solution is for a maximum or a minimum. Also, you are being asked for two answers (maximum and minumum) but have supplied only one solution. What is the answer to the other question?

 

[Karol]

L is divided as in:

a=the rectangle's side, b=the square's side
$$A=\left( \frac{L-3a}{2} \right)^2+\frac{\sqrt{3}}{4}a^2$$
$$A'=-3(L-3a)+\frac{\sqrt{3}}{4},~~A'=0:~a=\frac{3L}{2\sqrt{3}+9}$$
The answer should be:
$$a=\frac{3\sqrt{3}L}{4+3\sqrt{3}}$$

 

[kuruman]

a=the rectangle's side, b=the square's side

What rectangle and square? You have a square and an equilateral triangle.

Also, you did not answer the question about the minimum and maximum area cuts yet.

 

[Karol]

You are right, an equilateral triangle, not a rectangle.
As LCKurtz suggested, the maximum is at the end, where only the rectangle is, no triangle.

 

[kuruman]

In post #1 you found one value for $a$ assuming that the triangle is formed first. In post #4 you found another value for $a$ assuming that the square is formed first. How do these two values for $a$ can be used to answer the question posed by the problem? That's what @Ray Vickson tried to indicate to you in post #5.

 

[Karol]

I don't think it matters which i do first, the triangle or the square, but i have mistakes in both.
I made a mistake in calculating the triangle first:
$$A=\left( \frac{L-4a}{2} \right)^2+\frac{\sqrt{3}}{4}a^2,~~A'=-2(L-4a)+\frac{\sqrt{3}}{2}$$
$$A'=0:~~a=\frac{4L}{\sqrt{3}+9}$$
$$A''=8+\frac{\sqrt{3}}{2}>0$$
So it's a minimum.
the square first:
$$A=b^2+\frac{\sqrt{3}}{4}\left( \frac{L-2b}{4} \right)^2,~~A'=2b+\frac{\sqrt{3}}{32}$$
$$A'=0:~~a=\frac{(16-\sqrt{3})L}{64-2\sqrt{3}}$$

 

[kuruman]

I made a mistake in calculating the triangle first:
$A=\left( \frac{L-4a}{2} \right)^2+\frac{\sqrt{3}}{4}a^2,~~A'=-2(L-4a)+\frac{\sqrt{3}}{2}$

If this is the triangle first, how much wire is left for the square? How many sides does the square have? What fraction of the leftover wire is the length of one side?
Your square first answer is fraught with similar problems.
Try being systematic. Fill in the blanks
Square first
Let a = side of the square
S1. The amount of wire left after forming the square is ___________
S2. Therefore one side of the triangle is ___________ in length.

Triangle first
Let a = side of the triangle
T1. The amount of wire left after forming the triangle is ___________
T2. Therefore one side of the square is ___________ in length.

Once you fill in the blanks correctly, you can go ahead and calculate the areas in the two cases.

 

[Karol]

Triangle first
Let a = side of the triangle
$$A=\left( \frac{L-3a}{4} \right)^2+\frac{\sqrt{3}}{4}a^2,~~a=\frac{3L}{4\sqrt{3}+9}$$

Square first
Let b = side of the square
$$A=b^2+\frac{\sqrt{3}}{4}\left( \frac{L-4b}{3} \right)^2,~~b=\frac{\sqrt{3}L}{4(\sqrt{3}-9)}$$
$$a=\frac{L-4b}{3}=-\frac{3L}{\sqrt{3}-9}$$

 

[kuruman]

The expression you got for $b$ looks incorrect. You have $(\sqrt{3}-9)$ in the denominator which is a negative number.

 

[Karol]

none of the a's, nor the one in the triangle first method and nor the square first are correct. and they should be equal.
$$A=b^2+\frac{\sqrt{3}}{4}\left( \frac{L-4b}{3} \right)^2,~~b=\frac{\sqrt{3}L}{4(\sqrt{3}-9)}$$
$$a=\frac{L-4b}{3}=\frac{3L}{4\sqrt{3}+9}$$
But still the results aren't equal to the answer

 

[kuruman]

But still the results aren't equal to the answer

It may be because in the previous post your expression for $a$ is correct, but your expression for $b$ is not.

 

[Karol]

It may be because in the previous post your expression for a is correct, but your expression for b is not.

Which post, #14? only in this post i calculated b.
I found, in #12 and #14, the same value for a, the triangle's edge. that's good, because i think it doesn't matter which i calculate first.
But this a isn't equal to the answer the book gave for a.
The answer for b, the square's side, should be:
$$b=\frac{4L}{4+3\sqrt{3}}$$
And i didn't reach this answer either

 

[kuruman]

The answer for b, the square's side, should be:
$$b=\frac{4L}{4+3\sqrt{3}}$$

Note that 1/4 of the $b$ given above is $$\frac{L}{4+3\sqrt{3}}=\frac{\sqrt{3}L}{9+4\sqrt{3}}=a$$
It looks like what you think is the side is actually the perimeter of the square.

 

[Karol]

O.k., so the answers in the book are the perimeters of the triangle and the square.
But it's inverse, the given square's perimeter is 4 times my triangle's side, which i marked with a.
I found this same a in both my methods: the triangle first and the square first.

 

[kuruman]

So you found that when you form the triangle first, then the side should be the same as when you form the square first. Does that make sense?

 

[Karol]

Of course it makes sense. the rope can only be divided in one way between the triangle and the square in order to form the minimum, and the same for the maximum area.
In forming the triangle first i used the side of the triangle as the independent variable and the same for the square first. i expressed these sides as a function of L.
I assumed L is divided into 7 parts, 4 equal ones are b, the squares side and 3 equal ones are the triangles sides. the division is the same no matter which i express in terms of the other.

 

[Karol]

Did you see my respond?

 

[kuruman]

Did you see my respond?

Yes.

 

[Karol]

So? you didn't answer,

 

[kuruman]

What you said is OK, not how I would have said it, but you got the idea. Now ou have to figure out if you have a maximum or a minimum.

 

[Karol]

As LUKurtz said in post #3, the maximum (or minimum) can be at the endpoints.
The second derivative here is positive everywhere, and i didn't find 2 values for the first derivative, so the single, wrong, value if fund is the minimum.
At the endpoints (a=the triangle's side) :
At the right endpoint, only the triangle exists:
$$A=\left( \frac{L-3a}{4} \right)^2+\frac{\sqrt{3}}{4}a^2~\rightarrow~a=L/3: ~A=\frac{\sqrt{3}}{4}\left( \frac{L}{3} \right)=\frac{1}{12\sqrt{3}}L^2=0.05L^2$$
At the left endpoint a=0 and the whole area is of a square:
$$A=\left( \frac{L}{4} \right)^2=0.06L^2$$
The last is bigger so it's the maximum.
This result,according to the book, is right. it says to use it all for the square.
But still the minimum a isn't correct

 

[Mark44]

@Karol, your area function is quadratic in a, so the graph of this function is a parabola that opens upward. The domain is limited, though, since $0 \le a \le L$. Given that the graph is a parabola, there will be only one point at which the derivative is zero. Keep in mind though, that an extreme point can occur at these three places:

1. At points where the derivative is zero.
2. At endpoints of the domain.
3. At points where the function is defined, but the derivative is not defined.

 

[Karol]

Mark maybe you wrote your answer after i posted my last post, i used it there

 

[Mark44]

Mark maybe you wrote your answer after i posted my last post, i used it there

No, I saw your answer. You said your answer for a wasn't right, according to the book. Recognizing that the graph is a parabola on a limited domain should help you figure out what a needs to be.

 

[Karol]

My a is 0<a<L so it's location is correct
The answer in the book for a is $~\displaystyle~a=\frac{3\sqrt{3}L}{4+3\sqrt{3}}$
I have to multiply my answer for a by 3: $~\displaystyle~a=\frac{9L}{4\sqrt{3}+9}~$ since i was asked how to divide L, not a itself.
Still my answer isn't correct

 

[kuruman]

The answer in the book for a is $a=\frac{3\sqrt{3}L}{4+3\sqrt{3}}$
I have to multiply my answer for a by 3: $a=\frac{9L}{4\sqrt{3}+9}~$ since i was asked how to divide L, not a itself.
Still my answer isn't correct

$$a=\frac{3\sqrt{3}L}{4+3\sqrt{3}}=\frac{3\sqrt{3}\sqrt{3}L}{(4+3\sqrt{3})\sqrt{3}}=\frac{9L}{4\sqrt{3}+9}~!$$

 

[Karol]

And from my answer to the book's, without the L:
$$a=\frac{9}{4\sqrt{3}+9}=\frac{9}{\sqrt{3}(4+3\sqrt{3})}=\frac{3\sqrt{3}\sqrt{3}}{\sqrt{3}(4+3\sqrt{3})}=\frac{9}{4+3\sqrt{3}}$$
Thank you kuruman, Mark, Ray and LCKurtz