MHB A text problem on the study of extrema of a function

[urugvai]

Consider squares inscribed in different isosceles triangles with sides equal to 1. (One side of the square lies on the bottom.) Find the side of the largest squareView attachment 6546

 

[Evgeny.Makarov]

Hi, and welcome to the forum!

You can do this either with or without trigonometry. But in either case I have not found a formula for the largest square side, only an approximate value.

Let the $AC=b$ and $BH=h$. Then the area of $\triangle ABC$ is $(1/2)bh$. It also equals the sum of areas of $\triangle KBL$, the area of the square and the combined area of $\triangle AKN$ and $\triangle CLM$. Let the square side be $x$. Then the combined area of $\triangle AKN$ and $\triangle CLM$ is $(1/2)x(b-x)$. Can you express the area of the remaining figures and write the equation?

For the future, please read the http://mathhelpboards.com/rules/, especially rule #11 (click the Expand button on top).

 

[urugvai]

Hi, and welcome to the forum!

You can do this either with or without trigonometry. But in either case I have not found a formula for the largest square side, only an approximate value.

Let the $AC=b$ and $BH=h$. Then the area of $\triangle ABC$ is $(1/2)bh$. It also equals the sum of areas of $\triangle KBL$, the area of the square and the combined area of $\triangle AKN$ and $\triangle CLM$. Let the square side be $x$. Then the combined area of $\triangle AKN$ and $\triangle CLM$ is $(1/2)x(b-x)$. Can you express the area of the remaining figures and write the equation?

For the future, please read the http://mathhelpboards.com/rules/, especially rule #11 (click the Expand button on top).

Thank you very much, but now the solution implies the derivation of the formula for the dependence of the area on some value (which should be chosen), the derivative, critical points, and so on. And the proposed answer introduced me into to a grinding halt.
https://www.physicsforums.com/attachments/6547._xfImport

 

[Evgeny.Makarov]

now the solution implies the derivation of the formula for the dependence of the area on some value (which should be chosen)

We may choose $b$ as the parameter. It varies between 0 and 2. Alternatively, we can choose $h$, which varies between 0 and 1.

the derivative, critical points, and so on.

Yes, this has to be done, but you should show your work or describe your difficulties.

And the proposed answer introduced me into to a grinding halt.

This value seems correct, but deriving it analytically will require some work.

 

[MarkFL]

I think I would use coordinate geometry, where $\overline{AC}$ lies along the $x$-axis and $H$ is at the origin. We then see that point $L$ must lie along the line:

$$y=2x$$

Now, if the parameter $0<b<1$ corresponds to the length of $\overline{BH}$, then side $\overline{BC}$ will lie along the line:

$$\frac{x}{\sqrt{1-b^2}}+\frac{y}{b}=1$$

Since $$x=\frac{y}{2}$$, we have:

$$\frac{y}{2\sqrt{1-b^2}}+\frac{y}{b}=1$$

Solving for $y$, we obtain:

$$y\left(\frac{1}{2\sqrt{1-b^2}}+\frac{1}{b}\right)=1$$

$$y\left(\frac{b+2\sqrt{1-b^2}}{2b\sqrt{1-b^2}}\right)=1$$

$$y=\frac{2b\sqrt{1-b^2}}{b+2\sqrt{1-b^2}}$$

Applying the calculus, we indeed do find:

$$y_{\max}=\frac{2}{\sqrt{\left(\sqrt[3]{4}+1\right)^3}}\approx0.4805453399659586$$

I left those steps out for now to give you a chance to fill them in. :D

 

[urugvai]

$$\frac{x}{\sqrt{1-b^2}}+\frac{y}{b}=1$$

Thanks, but there was a difficulty, is this the sum of vectors?

 

[MarkFL]

Thanks, but there was a difficulty, is this the sum of vectors?

I used the two-intercept form of a line, and the fact that the distance between the two intercepts must be 1. :D

 

[urugvai]

I used the two-intercept form of a line, and the fact that the distance between the two intercepts must be 1. :D

The temperature of the brain in the head only increases :(

 

[MarkFL]

The temperature of the brain in the head only increases :(

A line having $x$-intercept at $(a,0)$ and $y$-intercept at $(0,b)$ will be given by:

$$\frac{x}{a}+\frac{y}{b}=1$$

Now, if the distance between these intercepts is 1, then we may write:

$$\sqrt{(a-0)^2+(0-b)^2}=1$$

This implies:

$$a^2=1-b^2$$

If we assume $0<a$, then we have:

$$a=\sqrt{1-b^2}$$

And so we obtain the line:

$$\frac{x}{\sqrt{1-b^2}}+\frac{y}{b}=1$$

 

[urugvai]

A line having $x$-intercept at $(a,0)$ and $y$-intercept at $(0,b)$ will be given by:

$$\frac{x}{a}+\frac{y}{b}=1$$

Now, if the distance between these intercepts is 1, then we may write:

$$\sqrt{(a-0)^2+(0-b)^2}=1$$

This implies:

$$a^2=1-b^2$$

If we assume $0<a$, then we have:

$$a=\sqrt{1-b^2}$$And so we obtain the line:

$$\frac{x}{\sqrt{1-b^2}}+\frac{y}{b}=1$$

Thank you for explaining this in more detail. Sorry that you need to spend so much time explaining
Two more questions
X is the half of the side of the square
Y is the side of the square?

 

[MarkFL]

Thank you for explaining this in more detail. Sorry that you need to spend so much time explaining

Glad to help, and I don't mind one bit explaining anything I've posted that's not clear. :D

Two more questions
X is the half of the side of the square
Y is the side of the square?

Yes, that's correct. (Yes)

 

[MarkFL]

This is what I would do to finish the problem:

We have that:

$$y=\frac{2b\sqrt{1-b^2}}{b+2\sqrt{1-b^2}}$$

Demonstrate that differentiating, we obtain:

$$y'=\frac{2\left(2\left(1-b^2\right)^{\frac{3}{2}}-b^3\right)}{\sqrt{1-b^2}\left(2\sqrt{1-b^2}+b\right)^2}$$

State with reasons why the only critical value(s) will come from:

$$2\left(1-b^2\right)^{\frac{3}{2}}-b^3=0$$

Demonstrate that this has the single real root:

$$b=\sqrt{\frac{1}{5}\left(4-2\sqrt[3]{2}+\sqrt[3]{4}\right)}$$

Demonstrate using the first derivative test that this critical value is at the global maximum.

And so we find:

$$y_{\max}=y\left(\sqrt{\frac{1}{5}\left(4-2\sqrt[3]{2}+\sqrt[3]{4}\right)}\right)=\frac{2}{\sqrt{\left(\sqrt[3]{4}+1\right)^3}}\approx0.4805453399659586$$