# Proof regarding a norman window max area...

• Jun 28th 2010, 02:10 PM
jkrowling
Proof regarding a norman window max area...
A norman window where the semicircle's radius is r (thus the base is 2r) and the height of the rectangle is h. Perimeter is constant, $\displaystyle P$. Show the values r and h are both $\displaystyle P/(pi+4)$ when the area is a maxiumum.

Please use a method not involving derivatives.
• Jun 28th 2010, 02:46 PM
skeeter
Quote:

Originally Posted by jkrowling
A norman window where the semicircle's radius is r (thus the base is 2r) and the height of the rectangle is h. Perimeter is constant, $\displaystyle P$. Show the values r and h are both $\displaystyle P/(pi+4)$ when the area is a maxiumum.

Please use a method not involving derivatives.

$\displaystyle P = 2h + 2r + \pi r$

and ...

$\displaystyle \displaystyle A = 2rh + \frac{\pi r^2}{2}$

solve for $\displaystyle h$ in the perimeter equation and substitute the result for $\displaystyle h$ in the area equation ... you should get area as a quadratic function of $\displaystyle r$. locate the vertex (max) using $\displaystyle \displaystyle r = \frac{-b}{2a}$.

let's see what you can do.
• Jun 28th 2010, 03:02 PM
jkrowling
OK...
$\displaystyle h=(P-2r-pi*r)/2$

$\displaystyle A=(2r*(P-2r-pi*r)/2)+(pi*r^2)/2$
$\displaystyle A(r)=r(P-2r-pi*r)+(pi*r^2)/2$

$\displaystyle r=-b/2a$
$\displaystyle r=-(P-2r-pi*r)/(2(1/2)*pi)$

I get to here, but I have trouble further simplifying.
• Jun 28th 2010, 03:33 PM
skeeter
$\displaystyle A(r) = 2r \cdot \frac{P - (2+\pi)r}{2} + \frac{\pi r^2}{2}$

$\displaystyle A(r) = Pr - (2+\pi)r^2 + \frac{\pi r^2}{2}$

$\displaystyle A(r) = Pr + \left(\frac{\pi}{2} - 2 - \pi\right)r^2$

$\displaystyle A(r) = Pr - \left(\frac{\pi}{2} + 2\right)r^2$

try again
• Jun 28th 2010, 04:39 PM
jkrowling
ok...

$\displaystyle A(r)=Pr-(pi/2+2)r^2$

$\displaystyle rmax=-b/2a$
$\displaystyle =-P/-2(pi/2+2)$
$\displaystyle =P/(pi+4)$

So then, to prove that this is also the max value for h, I assume I should plug the result into the original equation for h...?

$\displaystyle h=P-2(P/(pi+4))+pi(P/(pi+4))$

When I simplify this, I get

$\displaystyle h=(P*pi+4P-2P+P*pi)/(pi+4) * (1/2)$
$\displaystyle =(P*pi-p)/(pi+4)$

not the $\displaystyle P/(pi+4)$ that I was hoping for...??
• Jun 28th 2010, 05:41 PM
skeeter
$\displaystyle h = \frac{P - (2+\pi)r}{2}$

$\displaystyle h = \frac{P}{2} - \frac{2+\pi}{2} \cdot \frac{P}{\pi+4}$

common denominator ...

$\displaystyle h = \frac{P(\pi+4)}{2(\pi+4)} - \frac{P(2+\pi)}{2(\pi+4)}$

combine fractions ...

$\displaystyle h = \frac{P\pi + 4P - 2P - P\pi}{2(\pi+4)}$

combine like terms ...

$\displaystyle h = \frac{2P}{2(\pi+4)}$

simplify ...

$\displaystyle h = \frac{P}{\pi+4}$

you need to work on your algebra.