# Maximise a rectangle inside a semi-circle.

• Jun 3rd 2010, 04:16 AM
Bushy
Maximise a rectangle inside a semi-circle.
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 $\frac{1}{2}(\theta-\sin\theta)r^2$

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

I need to find where $A' = 0$

and I also think $\tan\frac{\theta}{2} = \frac{y}{2x}$

I have all these relationships yet I can't reduce it to one variable.
• Jun 3rd 2010, 05:07 AM
sa-ri-ga-ma
$r^2 = x^2 + (\frac{y}{2})^2$

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.
• Jun 3rd 2010, 05:16 AM
ebaines
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:

$
R^2 = x^2 + y^2
$

hence:

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

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

So now you need to determine for what value of x is it true that $\frac {dA} {dx}=0$, and then check that you have found a maximum, not a minimum. Can you take it from here?
• Jun 3rd 2010, 12:57 PM
Bushy
Thanks all, after making $A'=0$ i get $x=\frac{3}{\sqrt{2}}$

I can finish it from here.
• Jun 3rd 2010, 01:13 PM
ebaines
Quote:

Originally Posted by Bushy
Thanks all, after making $A'=0$ i get $x=\frac{3}{\sqrt{2}}$

I can finish it from here.

Better check your math .. I think you have an error..
• Jun 3rd 2010, 01:52 PM
Quote:

Originally Posted by Bushy
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 $\frac{1}{2}(\theta-\sin\theta)r^2$

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

I need to find where $A' = 0$

and I also think $\tan\frac{\theta}{2} = \frac{y}{2x}$

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.

$Sin\theta=\frac{y}{R}, Cos\theta=\frac{x}{R}$

$xy\ =\ area\ of\ half-rectangle\ =\ R^2Sin\theta\ Cos\theta$

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.

$2SinACosB=Sin(A+B)+Sin(A-B)$

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

Hence find $\frac{d}{d\theta}Sin2\theta=0$

$2Cos2\theta=0\ \Rightarrow\ Cos2\theta=0\ \Rightarrow\ 2\theta=\frac{{\pi}}{2}$

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