Calc optimization

First, let me say that this is a beautiful calc problem.

You should start by thinking about the wire and what you will want to call your solution.

Suppose you have the wire stretched out and you mark a point H on it and you say that you are going to cut it on H. Then the length of the first wire will be H and the length of the second wire will be (60-H)

Then you need to make a decision. What will you do with each wire? I say that you use the wire of length H to build the equilateral triangle.

If an equilateral triangle is made of wire length H then each side must be H/3. So now you need to find a formula for the area of said triangle. You do that by remembering that the base is H/3 and that you can find the height by using pythagoras theorem after splitting the triangle in two. Getting a right angled triangle with base H/6 and a hypotenuse of H/3.

Then you use base times height over 2 for that area.

Then for the circle you need to think again. If you are going to make a circle with a wire of length 60 - H then 60 - H is the circumference of the circle, but you need a formula for the area, which means you need a formula for the radius.

Fortunately, Circumference over diameter equals pi So then you can plug 60 - H as your circumference and then solve for the diameter and then divide that by two.

Then the area of the circle is pi times the radus squared.

Then you add both of the formulas for the areas into a nice f(H) function and differentiate.

Thank you. I made the mistake of making h the circle. Now my model is ugly as shit.

It turns out ugly as shit either way, I think.

desmos.com/calculator/ekvu5stsdo

Fuck, desmos.com/calculator/1z37k3ibey

Not really. The answer is simply the length of the wire divided by (1+the ratio between the areas of a circle and equilateral triangle with equal diameter)

*equal circumference

Yeah but the way i did it comes off ugly right off the bat cuz it used equilateral formula so ((3)^(1/2)/4)(60-l)^2 or something then the circle. Just hell. Thanks tons man. Im off to do this

>can't even calculate the areas properly
what are you doing here

He's trying to fuck up brainlet OP.

Let [math]1_1[/math] and [math]l_2[/math] be the perimeters of the circle and the equilateral triangle, respectively. We have
[eqn]
l_1 + l_2 = 60.
[/eqn]
Rearranging, this becomes
[eqn]
l_2 = 60 - l_1.
[/eqn]

Similarly, let [math]A_1[/math] and [math]A_2[/math] be the areas of each shape, respectively, and set [math]A = A_1 + A_2[/math].

We want to maximize and minimize [math]A[/math]. First, we maximize.

Recalling the formulae for the circumference and area of a circle, we have
[eqn]
l_1 = 2\pi r
\Rightarrow r = \frac{l_1}{2\pi},\\
A_1 = \pi r^2
= \pi\left(\frac{l_1}{2\pi}\right)^2
= \frac{l_1^2}{4\pi}.
[/eqn]

Recalling the formula for the area of a triangle, we have
[eqn]
A_2 = \frac{bh}{2}
[/eqn]
where
[eqn]
b = \frac{l_2}{3},\\
h = \frac{l_2}{3}\sin\left(\frac{\pi}{3}\right).
[/eqn]
That is,
[eqn]
A_2 = \frac{\sqrt{3}}{36}l_2^2
= \frac{\sqrt{3}}{36}(60 - l_1)^2
= \frac{\sqrt{3}}{36}(3600 + l_1^2 - 120l_1).
[/eqn]

We have
[eqn]
A = A_1 + A_2
= \frac{l_1^2}{4\pi} + \frac{\sqrt{3}}{36}(3600 + l_1^2 - 120l_1)\\
\Rightarrow \frac{\mathrm{d}}{\mathrm{d}l_1}[A]
= \frac{l_1}{2\pi} + \frac{\sqrt{3}}{36}(2l_1 - 120)\\
\Rightarrow \frac{\mathrm{d}^2}{\mathrm{d}l_1^2}[A]
= \frac{1}{2\pi} + \frac{\sqrt{3}}{18}
[/eqn]

Since its second derivate is strictly positive, we know that the function [math]A(l_1)[/math] is accelerating; some classes call this function "concave up."

This tells us that the maximum is located on the boundary of the interval, so [math]l_1[/math] is either 0 or 60.

We have
[eqn]
A(0) = A_2(0) = 100\sqrt{3}\\
A(60) = A_1(60) = \frac{900}{\pi}
[/eqn]

Since the area is maximized for a value of 60, we can see that the wire should not be cut, and instead the full length should be used to construct a circle.

Next, we minimize. We set the first derivate equal to 0 and obtain
[eqn]
0 = l_1\left(\frac{1}{2\pi} + \frac{\sqrt{3}}{18}\right) - \frac{10\sqrt{3}}{3}
= l_1\frac{18 + 2\pi\sqrt{3}}{36\pi} - \frac{10\sqrt{3}}{3}\\
\Rightarrow l_1^* = \frac{120\pi\sqrt{3}}{18 + 2\pi\sqrt{3}}
[/eqn]

This value is approximately 22.607, which is on our interval; i.e. it is between the values 0 and 60. Indeed, you can confirm that plugging in the value to [math]A(l_1)[/math] will yield a value lower than [math]A(0)[/math] and [math]A(60)[/math].

We cut the wire into two pieces, say the first of which has length [math]l_1^*[/math]. We use that first piece to form the circle, and the second piece to to form the equilateral triangle.