Respuesta :
Answer:
Step-by-step explanation:
Given Perimeter of window
[tex]P=16\ ft[/tex]
Suppose x and y be the dimension of rectangular window
Perimeter of rectangular portion [tex]=x\cdot y[/tex]
Perimeter of semicircle[tex]=\pi (0.5x)[/tex]
so
[tex]2(x+y)+0.5\pi x=16[/tex]
[tex]y=8-0.25\pi x-x[/tex]
Area of the window is
[tex]A=xy+\frac{\pi x^2}{4} [/tex]
[tex]A=x(8-0.25\pi x-x)+0.25\pi x^2[/tex]
differentiate A w.r.t x to get maximum or minimum value
[tex]\frac{\mathrm{d} A}{\mathrm{d} x}=8-(0.5\pi +2)x+\frac{\pi x}{2}[/tex]
put [tex]\frac{\mathrm{d} A}{\mathrm{d} x}=0[/tex] to get maxima/minima
thus [tex]x=4 ft[/tex]
for [tex]x=4[/tex] , [tex]y=8-0.25\pi \cdot 4-4[/tex]
[tex]y=0.858 ft[/tex]
The maximum area of the window has a semicircle with diameter 4.48 ft and the length of the rectangle is 2.24 ft.
Let x represent the radius of the semicircle, hence the width of the rectangle = 2x. Also, the length of the rectangle is y
The perimeter of the window = πx + 2y + 2x = x(π + 2) + 2y
16 = x(π + 2) + 2y
[tex]y=\frac{16-x(\pi +2)}{2}[/tex]
The area of the window (A) = 0.5πx² + 2yx[tex]A=\frac{1}{2}\pi x^2 +2x(\frac{16-x(\pi + 2)}{2} )\\\\A=\frac{1}{2}\pi x^2 +16x-\pi x^2-2x^2\\\\A=16x-x^2(2+\frac{\pi}{2} )\\\\The \ maximum\ value\ of\ x\ is\ at\ dA/dx=0\\\\A'=16-2x(2+\frac{\pi}{2} )\\\\2x(2+\frac{\pi}{2} )=16\\\\x(2+\frac{\pi}{2} )=8\\\\x=2.24\ ft[/tex]
[tex]y=\frac{16-x(\pi +2)}{2}=\frac{16-2.24(\pi +2)}{2}=2.24[/tex]
Hence the maximum area of the window has a semicircle with diameter 4.48 ft and the length of the rectangle is 2.24 ft.
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