Finding the minimum area of a triangle by using the derivative of a function

[multicube]

There is a rectangular piece of paper with a length of 23.2 cm and height of 22 cm. The lower right corner is folded to the top edge forming a triangle. Find the minimum area of a triangle that can be constructed.


At first, I found the relationship between the base and height of the triangle, and then I differentiated the area, and then the result can be shown here, as a graph.

(x^2/2)sqrt{11/(x-11)} - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

and then this wasn't the answer, so, I thought that the edge of the triangle could perfectly meet with the edge of the rectangle and found another answer.

but this was also incorrect.
And this problem specifically mentions that a 'triangle' must be constructed, so my attempts to find a square failed, too.
I am very lost, and could you please tell me how to get back on track?

 

[MarkFL]

Hello, and welcome to FMH!

I would label some sides and angles as follows:



By Pythagoras, we find:

1.) [MATH]a^2+b^2=c^2[/MATH]
2.) [MATH](y-a)^2+d^2=a^2[/MATH]
Now, by similarity, we may state:

[MATH]\frac{b}{y}=\frac{a}{d}[/MATH]
Or:

[MATH]b=\frac{y}{\sqrt{y(2a-y)}}a=\sqrt{\frac{y}{2a-y}}a[/MATH]
And so the area of the triangle is:

[MATH]A=\frac{1}{2}ab=\frac{1}{2}\sqrt{\frac{y}{2a-y}}a^2[/MATH]
Can you state the domain of this function? Plotting this area function, we find it looks like this:



We now have the area of the triangle in terms of 1 variable (recall that \(y\) is a parameter and can be treated as a constant). See if you can compute \(A'(a)\) and equate it to zero to find the value of \(a\) that minimizes the area of the triangle.

 

[MarkFL]

I would likely square and arrange as:

[MATH]\frac{4A^2}{y}=\frac{a^4}{2a-y}[/MATH]
We have a constant times the square of the area on the left, and minimizing the product of a constant and the square of the area is the same as minimizing the area itself with regards to finding the critical value(s), and now the differentiating is somewhat simpler:

[MATH]\frac{d}{da}\left(\frac{4A^2}{y}\right)=\frac{(2a-y)4a^3-a^4(2)}{(2a-y)^2}=\frac{2a^3(2(2a-y)-a)}{(2a-y)^2}=\frac{2a^3(3a-2y)}{(2a-y)^2}=0[/MATH]
Observing we must have:

[MATH]\frac{y}{2}<a\le y[/MATH]
The only root we need consider is:

[MATH]3a-2y=0\implies a=\frac{2}{3}y[/MATH]
We can see that on [MATH]\left(\frac{y}{2},\frac{2y}{3}\right)[/MATH] we have:

[MATH]\frac{d}{da}\left(\frac{4A^2}{y}\right)<0[/MATH]
And on [MATH]\left(\frac{2y}{3},y\right][/MATH] we have:

[MATH]\frac{d}{da}\left(\frac{4A^2}{y}\right)>0[/MATH]
And so we may conclude:

[MATH]A_{\min}=A\left(\frac{2}{3}y\right)=\frac{2\sqrt{3}}{9}y^2[/MATH]
Here's a live graph that will help you explore the problem:

Desmos | Graphing Calculator

www.desmos.com

 

[multicube]

Hey-I found the answer, thanks to your help!!!
There was something I missed when I posted the wolframalpha function-the domain of the function.
I forgot to consider what quantity a is when b is 23.2, because a should always be bigger than that.
The picture below is what the A looks like, and a should always be bigger than that purple line in the picture.
Thank you very much for setting me back on track-this question had been bothering me for three days!

 

[MarkFL]

I recall encountering a variant of this problem many years ago, and that was to minimize the length of the crease \(c\).

We may state:

[MATH]c^2=a^2+b^2=a^2+\frac{y}{2a-y}a^2=a^2\left(1+\frac{y}{2a-y}\right)=\frac{2a^3}{2a-y}[/MATH]
[MATH]\frac{d}{da}\left(\frac{c^2}{2}\right)=\frac{(2a-y)3a^2-a^3(2)}{(2a-y)^2}=\frac{a^2(4a-3y)}{(2a-y)^2}=0[/MATH]
From this we conclude:

[MATH]a=\frac{3}{4}y[/MATH]
The first-derivative test show this critical value is at a minimum, and so we may conclude that:

[MATH]c_{\min}=\sqrt{\frac{2\left(\frac{3}{4}y\right)^3}{2\left(\frac{3}{4}y\right)-y}}=\frac{3\sqrt{3}}{4}y[/MATH]

 

[MarkFL]

Here's a better image:


And a better live graph:

Desmos | Graphing Calculator

www.desmos.com