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Calculus Optimization Problem

drasord

New member
Dec 1, 2013
5
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

Administrator
Staff member
Feb 24, 2012
13,775
Can you show us what you have tried so far so we know where you are stuck and can best offer help?
 

drasord

New member
Dec 1, 2013
5
Absolutely - sorry for the delay! This is my understanding of how to "optimize" the problem:

optimization.png

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

Administrator
Staff member
Feb 24, 2012
13,775
You have found the correct critical value:

\(\displaystyle x=\frac{R}{\sqrt{2}}\)

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

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

drasord

New member
Dec 1, 2013
5
alw.png

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

A = \(\displaystyle 2x * sqrt(R^2 - x^2)\)
 
Last edited:

drasord

New member
Dec 1, 2013
5
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

Administrator
Staff member
Feb 24, 2012
13,775
The base of the rectangle is:

\(\displaystyle 2x=2\cdot\frac{R}{\sqrt{2}}=\sqrt{2}R\)

The height is:

\(\displaystyle y=\sqrt{R^2-\frac{R^2}{2}}=\frac{R}{\sqrt{2}}\)

Thus area = base times height:

\(\displaystyle A=\sqrt{2}R\cdot\frac{R}{\sqrt{2}}=R^2\)