Calculus 1 Homework help, PLEASE!?

Question

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1. A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 24 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.
Enter your answers correct to two decimal places. Remember to take into account in your calculations that there is only half of a circle. The length of the bottom of the rectangle is __ feet. The length of each of the sides of the rectangle is __ feet.

2. The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, how far in ft from the stronger source should an object be placed on the line between the sources so as to receive the least TOTAL illumination (the sum of the illumination from each source)

Answer 1

Very interesting. In 1.), let's say w is the width of the window. The perimiter of the semicircle is pi * w/2. The height of the window is thus found from

2h = 24 - w - pi * w/2
h = (24 - w - pi * w/2) / 2

The area of the window is the area of the rectangle, w * h, plus the area of the semicircle, pi * (w/2)^2 / 2.

A = w * h + pi * w^2 / 8

Now substitute the expression for h:

A = w * [(24 - w - pi * w/2) / 2] + pi * w^2 / 8
= 12w - w^2/2 - pi * w^2/8

Now take the derivative of A and set it equal to 0:

0 = 12 - w - pi * w/4

Answer 2

(1) If the height of rectangular portion of the window is H and the radius of the semicircular portion is R, then the total perimeter is 2H + 2R + piR. If this equals 24, we can solve for H in order to make a substitution:

2H + 2R + piR = 24
2H = -2R - piR + 24
H = -R - .5piR +12

OK, so we really want to maximize the area of the window if we are trying to maximize the light it lets through. The area of this shape is the rectangle's area plus the semicircle's area:

A = 2RH + .5piR^2
make the above substitution for H:
A = 2R(-R - .5piR + 12) + .5piR^2
A= -2R^2 -piR^2 + 24R + .5piR^2
A = -2R^2 - .5piR^2 + 24R

A' = -4R - piR + 24
find critical values:
0 = -4R - piR + 24
24/(4+pi) = R = 3.36

The width of the bottom of the rectangle is 2R = 48/(4+pi) = 6.72 and the height of the rectangle is -R -.5piR +12 = -3.36 - .5pi*3.36 + 12 = 3.36

Answer 3

W(1 + π/2) + 2L = 24
A = LW
L = (24 - W(1 + π/2))/2
A = (24W - W^2(1 + π/2))/2
dA/dW = 12 - W(1 + π/2)
W = 12/(1 + π/2) ft. ≈ 4.6678 ft.
L = (24 - 12(1 + π/2)/(1 + π/2))/2 = 6 ft.
A = 72/(1 + π/2) ft^2 ≈ 28.007 ft^2

L = L0(1/d^2 - 3/(10 - d)^2)
dL/dd = -2/d^3 - 6/(10 - d)^3
-2(10 - d)^3 = 6d^3
-2,000 + 600d - 6d^2 + 2d^3 = 6d^3
2d^3 - 3d^2 + 300d - 1,000 = 0
d ≈ 3.215136 ft from the dimmer source.