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?

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

The first and second derivatives are,

Area, A(r)

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)

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)