Respuesta :
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
[tex]=h+h+2r+\pi r=2h+2r+\pi r=12[/tex]
The area of the window is what we want to maximize.
Area of the Window=Area of Rectangle+Area of Semicircle
[tex]=2hr+\frac{\pi r^2}{2}[/tex]
We are trying to Maximize A subject to [tex]2h+2r+\pi r=12[/tex]
[tex]2h+2r+\pi r=12\\h=6-r-\frac{\pi r}{2}[/tex]
The first and second derivatives are,
Area, A(r)[tex]=2r(6-r-\frac{\pi r}{2})+\frac{\pi r^2}{2}}=12r-2r^2-\frac{\pi r^2}{2}[/tex]
Taking the first and second derivatives
[tex]A'\left( r \right) = 12 - r\left( {4 + \pi } \right)\\A''\left( r \right) = - 4 - \pi[/tex]
From the two derivatives above, we see that the only critical point of r
[tex]A'\left( r \right) = 12 - r\left( {4 + \pi } \right)=0[/tex]
[tex]r = \frac{{12}}{{4 + \pi }} = 1.6803[/tex]
Since the second derivative is a negative constant, the maximum area must occur at this point.
[tex]h=6-1.6803-\frac{\pi X1.6803}{2}=1.6803[/tex]
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)
