MHB What is the Optimal Rectangle in a Semicircle with Radius R?

[drasord]

I'm really stuck on this problem. Could anyone provide some help?

Find the length and width of the rectangle of largest area that can be inscribed in a semicircle of radius R, assuming that one side of the rectangle lies on the diameter of the semicircle. Also, find the area of this rectangle. Draw a neat diagram.

Thanks!

 

[MarkFL]

Can you show us what you have tried so far so we know where you are stuck and can best offer help?

 

[drasord]

Absolutely - sorry for the delay! This is my understanding of how to "optimize" the problem:

View attachment 1725

So I've found the area, I believe. But I need to find the "rectangle of largest area that can be inscribed in a semicircle of radius R". I'm confused about how to do this. And what does the professor mean by "a neat diagram"?

 

[MarkFL]

You have found the correct critical value:

$$x=\frac{R}{\sqrt{2}}$$

The base of the rectangle is $2x$. The height is $y$.

So, what is $$y\left(\frac{R}{\sqrt{2}} \right)$$ ?

 

[drasord]

View attachment 1727

So now I have A, length, and width. Correct?

A = $$2x * sqrt(R^2 - x^2)$$

 

[drasord]

Or now I have to find area:

A = l * w
A = 2R/sqrt(2) * sqrt(R^2/2)
A = sqrt(2) * sqrt(x) * sqrt(x^2)

?

 

[MarkFL]

The base of the rectangle is:

$$2x=2\cdot\frac{R}{\sqrt{2}}=\sqrt{2}R$$

The height is:

$$y=\sqrt{R^2-\frac{R^2}{2}}=\frac{R}{\sqrt{2}}$$

Thus area = base times height:

$$A=\sqrt{2}R\cdot\frac{R}{\sqrt{2}}=R^2$$