Respuesta :
Answer:
The dimensions that maximize the area of the window are:
Width: 8.4 ft
Height: 4.2 ft
Step-by-step explanation:
We have a Norman window, in which the diameter of the semicircle is equal to the width of the rectangle.
The perimeter is made of 2 heights of the rectangle, one width of the rectangle plus half a circumference.
Then, if w is the diameter of the semicircle (and the width of the rectangle) and h is the other side of the rectangle we can write the perimeter as:
[tex]P=2h+w+\dfrac{\pi\cdot w }{2}=30\\\\\\2h+(1+0.5\pi)w=30\\\\h=\dfrac{30-(1+0.5\pi)w}{2}=15-(0.5+0.25\pi)w[/tex]
With the restriction for the perimeter (P=30), we can express h in funtion of w.
Now, we can express the area only in function of w and maximize its value:
[tex]A=A_c/2+A_r=\dfrac{1}{2}\cdot\dfrac{\pi w^2}{4}+hw=\dfrac{\pi w^2}{8}+[15-(\dfrac{1}{2}+\dfrac{\pi}{4})w]\cdot w\\\\\\A=15w-(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi }{8})w^2=15w-(\dfrac{1}{2}+\dfrac{\pi }{8})w^2[/tex]
To maximize the area, we derive and equal to 0:
[tex]\dfrac{dA}{dw}=15(1)-(\dfrac{1}{2}+\dfrac{\pi }{8})(2w)=0\\\\\\15-(1+\pi/4)w=0\\\\\\w=\dfrac{15}{1+\pi/4}\approx8.4[/tex]
Now, we can calculate the height as:
[tex]h=15-(0.5+0.25\pi)w\\\\\\h\approx 15-(1.2854)\cdot 8.4=15-10.8=4.2[/tex]
