Question

g A window is being built and the bottom is a rectangle and the top is a semi-circle. If there is 12 meters of framing materials, what must the dimensions of the window be to let in the most light?

Answer 1

Answer:

Semicircle of radius of 1.6803 meters

Rectangle of dimensions 3.3606m x 1.6803m

Step-by-step explanation:

Let the radius of the semicircle on the top=r  

Let the height of the rectangle =h  

Since the semicircle is on top of the window, the width of the rectangular portion =Diameter of the Semicircle =2r

The Perimeter of the Window

=Length of the three sides on the rectangular portion + circumference of the semicircle

$=h+h+2r+\pi r=2h+2r+\pi r=12$

The area of the window is what we want to maximize.

Area of the Window=Area of Rectangle+Area of Semicircle

$=2hr+\frac{\pi r^2}{2}$

We are trying to Maximize A subject to $2h+2r+\pi r=12$

$2h+2r+\pi r=12\\h=6-r-\frac{\pi r}{2}$

The first and second derivatives are,

Area, A(r)$=2r(6-r-\frac{\pi r}{2})+\frac{\pi r^2}{2}}=12r-2r^2-\frac{\pi r^2}{2}$

Taking the first and second derivatives

$A'\left( r \right) = 12 - r\left( {4 + \pi } \right)\\A''\left( r \right) = - 4 - \pi$

From the two derivatives above, we see that the only critical point  of r

$A'\left( r \right) = 12 - r\left( {4 + \pi } \right)=0$

$r = \frac{{12}}{{4 + \pi }} = 1.6803$

Since the second derivative is a negative constant, the maximum area must occur at this point.

$h=6-1.6803-\frac{\pi X1.6803}{2}=1.6803$

So, for the maximum area the semicircle on top must have a radius of 1.6803 meters and the rectangle must have the dimensions 3.3606m x 1.6803m ( Recall, The other dimension of the window = 2r)