# Thread: Proof regarding a norman window max area...

1. ## 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, $P$. Show the values r and h are both $P/(pi+4)$ when the area is a maxiumum.

Please use a method not involving derivatives.

2. 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, $P$. Show the values r and h are both $P/(pi+4)$ when the area is a maxiumum.

Please use a method not involving derivatives.
$P = 2h + 2r + \pi r$

and ...

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

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

let's see what you can do.

3. OK...
$h=(P-2r-pi*r)/2$

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

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

I get to here, but I have trouble further simplifying.

4. $A(r) = 2r \cdot \frac{P - (2+\pi)r}{2} + \frac{\pi r^2}{2}$

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

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

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

try again

5. ok...

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

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

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...?

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

When I simplify this, I get

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

not the $P/(pi+4)$ that I was hoping for...??

6. $h = \frac{P - (2+\pi)r}{2}$

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

common denominator ...

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

combine fractions ...

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

combine like terms ...

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

simplify ...

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

you need to work on your algebra.