Calculus Optimization Problem

[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!

 

[Romsek]

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!

Put your rectangle in your semicircle. What are the points of it's 4 vertices? One of them is (-x,0). You find the other 3.

Given those 4 points what is the area of your rectangle. How would you go about finding the maximum of this expression?

Come back if you still need help.

 

[pka]

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.

Can you maximize \(\displaystyle A(t)=R^2\cos(t)\sin(t)~? \)

If so then the area you want is \(\displaystyle 2A(t_{max}). \)

 

[drasord]

This is what I have done so far. I don't think it is correct.

 

[drasord]

I'm really stuck on this one. This is what I've done today, coming directly from class notes:



So it's been "optimized". However, I need to find the length and width. So I know length = 2x and width = y. But I'm blanking on how to find these values with the work I've done. Is the absolute max useful?

 

[HallsofIvy]

It's a little bit hard to read your writing but I think you have found that \(\displaystyle x= R/\sqrt{2}\). Of course, \(\displaystyle y= \sqrt{R^2- x^2}\) so what is y?

You yourself said that "length= 2x" and "width= y".

 

[Romsek]

I'm really stuck on this one. This is what I've done today, coming directly from class notes:

View attachment 3479

So it's been "optimized". However, I need to find the length and width. So I know length = 2x and width = y. But I'm blanking on how to find these values with the work I've done. Is the absolute max useful?

I can't read any of that so I'll just say how to do it. You've obviously given it a go.

Put your rectangle into the circle of radius R. The two bottom points are (-x,0) and (x,0). I'm sure you got that.

the equation for a circle of radius R is x² + y² = R². You know the two upper vertices touch that circle so at that point you must have

y = sqrt(R² - x²) where you choose the positive square root.

so now you have your 4 vertices (-x,0), (-x, sqrt(R² - x²)), (x, sqrt(R² - x²)), and (x,0)

the length of this rectangle is 2x, the width is sqrt(R² - x²) so the area is A = l * w = 2x * sqrt(R² - x²)

in other words A(x) = (2x) sqrt(R²-x²)

This is just a function in x that you know how to maximize. Take it's derivative, set it equal to 0 and solve.

You get a very clean answer in terms of R.

 

[drasord]

It's a little bit hard to read your writing but I think you have found that \(\displaystyle x= R/\sqrt{2}\). Of course, \(\displaystyle y= \sqrt{R^2- x^2}\) so what is y?

You yourself said that "length= 2x" and "width= y".

So I get [sorry, this should be clearer]:



Correct?

 

[drasord]

And now to find Area:

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

?

 

[Romsek]

And now to find Area:

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

?

x = R / sqrt(2) is correct.

in the last line above where did the x's come from??

the 2nd line is correct

 

[HallsofIvy]

And now to find Area:

Actually, the problem did NOT ask you to find the area!
"Find the length and width ..."

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

Do you not know that \(\displaystyle \sqrt{R^2}= R\), since R must be positive? \(\displaystyle \sqrt{\frac{R^2}{2}}= \frac{R}{\sqrt{2}}\)

A = sqrt(2) * sqrt(x) * sqrt(x^2)

?

Where did the x in that last line come from?
You said \(\displaystyle A= \frac{2R}{\sqrt{2}}*\sqrt{\frac{R^2}{2}}= \frac{2R}{\sqrt{2}}*\frac{R}{\sqrt{2}}\)
Now do the arithmetic.