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loveyeti106838
04.04.2020 •
Mathematics
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?
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Ответ:
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
The area of the window is what we want to maximize.
Area of the Window=Area of Rectangle+Area of Semicircle
We are trying to Maximize A subject to![2h+2r+\pi r=12](/tpl/images/0582/0705/0318d.png)
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}](/tpl/images/0582/0705/c989c.png)
Taking the first and second derivatives
From the two derivatives above, we see that the only critical point of r
Since the second derivative is a negative constant, the maximum area must occur at this point.
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)
Ответ:
6 years later,
Amy is 23 years old
Diane is 46 years old
Now,
Amy, 17 years old
Diane, 40 years old.