Maximise a rectangle inside a semi-circle.

Aug 2009
69
5
Where the semicircle has radius of 3 and the rectangle with height x and width y touches the edges of the semicircle.

My thoughts are

Area (rectangle) = [semicircle] - [2 segments]

where the area a segment is \(\displaystyle \frac{1}{2}(\theta-\sin\theta)r^2\)

\(\displaystyle A = \frac{1}{2} \pi \times 3^2 - f(x,y,\theta)\)

I need to find where \(\displaystyle A' = 0\)

and I also think \(\displaystyle \tan\frac{\theta}{2} = \frac{y}{2x}\)

I have all these relationships yet I can't reduce it to one variable.
 
Jun 2009
806
275
\(\displaystyle r^2 = x^2 + (\frac{y}{2})^2\)

Area of the rectangle A = xy.

From the first equation find x in terms of r and y.
Put it in the second equation and find dA/dy. Equate it to zero.
 
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Jun 2008
1,389
513
Illinois
Whenever you are asked to "maximize" or "minimize" something, you want to develop an equation for that quantity that is in one variable, then take teh derivative with respect to that variable and set it to zero. Finally, check that you have a max, and not a min. Consider the semi-circle with center at (0,0), radius R, and a rectange inscibed inside of width 2x and height y. The equation of any point on the circle is given by:

\(\displaystyle
R^2 = x^2 + y^2
\)

hence:

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

And the area of the rectangle is
\(\displaystyle
A = 2xy = 2x \sqrt {R^2 - x^2}
\)

So now you need to determine for what value of x is it true that \(\displaystyle \frac {dA} {dx}=0\), and then check that you have found a maximum, not a minimum. Can you take it from here?
 
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Aug 2009
69
5
Thanks all, after making \(\displaystyle A'=0\) i get \(\displaystyle x=\frac{3}{\sqrt{2}}\)

I can finish it from here.
 
Jun 2008
1,389
513
Illinois
Thanks all, after making \(\displaystyle A'=0\) i get \(\displaystyle x=\frac{3}{\sqrt{2}}\)

I can finish it from here.
Better check your math .. I think you have an error..
 
Dec 2009
3,120
1,342
Where the semicircle has radius of 3 and the rectangle with height x and width y touches the edges of the semicircle.

My thoughts are

Area (rectangle) = [semicircle] - [2 segments]

where the area a segment is \(\displaystyle \frac{1}{2}(\theta-\sin\theta)r^2\)

\(\displaystyle A = \frac{1}{2} \pi \times 3^2 - f(x,y,\theta)\)

I need to find where \(\displaystyle A' = 0\)

and I also think \(\displaystyle \tan\frac{\theta}{2} = \frac{y}{2x}\)

I have all these relationships yet I can't reduce it to one variable.
You can use Pythagoras' theorem or trigonometry.

If x=half the length of the rectangle and y=height,
then the x giving max area for a half rectangle will give max area for the full rectangle.

So you can just work with a quadrant.

\(\displaystyle Sin\theta=\frac{y}{R}, Cos\theta=\frac{x}{R}\)

\(\displaystyle xy\ =\ area\ of\ half-rectangle\ =\ R^2Sin\theta\ Cos\theta\)

Minimum area is zero, so you can easily solve for maximum area.
Differentiate the area function with respect to the angle and equate to zero.

\(\displaystyle 2SinACosB=Sin(A+B)+Sin(A-B)\)

\(\displaystyle Sin\theta\ Cos\theta=0.5\left(Sin2\theta+Sin0\right)=0.5Sin2\theta\)

Hence find \(\displaystyle \frac{d}{d\theta}Sin2\theta=0\)

\(\displaystyle 2Cos2\theta=0\ \Rightarrow\ Cos2\theta=0\ \Rightarrow\ 2\theta=\frac{{\pi}}{2}\)


Maximum area of the rectangle is \(\displaystyle 2R^2(0.5)Sin\frac{{\pi}}{2}=R^2\)