Can someone solve this?

Question

A normal window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. If the perimeter of the window is 20ft, what dimensions will admit the most light (maximum area)

Answer 1

perimeter = 2 * l + w + (π * w / 2) = 20
area = l * w + (π * w² / 8)

Solving for l in the first equation gives you:

2 * l = 20 - w - (π * w / 2)
l = 10 - w/2 - (π * w / 4)

Substitute this into the area equation:

area = (10 - w/2 - (π * w / 4)) * w + (π * w² / 8)
area = 10w - w²/2 - π*w²/4 + π*w²/8

Now, take the first derivative of this, set it equal to zero, and solve for w:

10 - w - π*w/2 + π*w/4 = 0
40 - 4w - 2π*w + π*w = 0
4w + π*w = 40
w * (4 + π) = 40
w = 40 / (4 + π) ≈ 5.6009915351155737591047862548629
l = 10 - w/2 - (π * w / 4) ≈ 2.8004957675577868795523931274314

l & w are the dimensions of the rectangle, and w is the diameter of the semicircle above it.

Answer 2

Let w =width of window, h=height of window (rectangular part)
then perimeter of semicircle = pi w/2
total perimeter = 2L + w + pi w/2 =20ft
so 2L = 20 - w (1 + pi/2)
L = 10 - (w/2)(1 + pi/2)
Area = A(w) = wL + pi w^2/8
A(w) = pi w^2/8 + w(10 - (w/2)(1 + pi/2)
= w^2 (pi/8 -(1+pi/2)/2 ) + 10w
= -0.8927 w^2 + 10w
A(w) is max when A'(w) = 0
A'(w) = - 1.7854w + 10 = 0
or w = 10/1.7854 = 5.6

choose a width of w = 5.6ft

Answer 3

w= x l=y
(x+y= 10
s=xy +pi*x^2/8
so s= 1/8( 8xy+pix^2) and y = 10-x
s=1/4(80x-8x^2+pix^2
s´= 1/4( (2pi-16)x+80) =0 so x= 40/(8-pi) and y = 10-40/(8-pi)
all units in feet

Answer 4

the length and width of the rectangular part