1. If 400 ft of fence is to be used, which shape generates the larger area, a circle or a square?

SQUARE:

A = s^2

using 400 ft of fence, each side of the square should be

s = 100 ft

A = s^2

A = (100)^2

A = 10,000


CIRCLE:

C = 2 * pi * r

400 = 2 * pi * r

r = 400/(2 * pi )

r = 63.7

A = pi * r^2

A = pi * (63.7)^2

A = 12,740


Comparing the areas of square and circle,

CIRCLE ---> larger area

2. What are the dimensions of a triangle of maximum area that can be inscribed in a circle such that one side of the triangle passes through the center of the circle.

solution:

let

b = base of triangle

h = height of triangle

t = angle between the hypotenuse C and the height h

C = hypotenuse = 2r



A = 1/2 b * h ---> equation1

h = C * cos t

b = C * sin t

substituting b and h in equation1

A = (1/2) C^2 cos t sin t ---> equation2


from double angle formulas:

sin(2t) = 2 sin t cos t

cos t sin t = (1/2)sin(2t) ---> equation3


substituting equation3 in equation2

A = (1/4) C^2 sin (2t)


using the double angle form,

A = (1/4) C^2 (2 cos t sin t)


derivative of A with respect to t ---> derivative of a product

dA/dt = (1/4) C^2 (2 cos t sin t)

dA/dt = (1/4) C^2 [ 2 (cos t cos - sin t sint) ]

dA/dt = (1/4) C^2 [ 2 (cos^2 t - sin^2 t) ]


from double angle formulas:

cos^2 t - sin^2 t = cos 2t


substituting

dA/dt = (1/4) C^2 [ 2 (cos 2t) ]

dA/dt = (1/4) C^2 * 2 cos 2t


to find the maximum area, equate dA/dt = 0

dA/dt = (1/4) C^2 * 2 cos (2t) = 0


as the angle between the hypotenuse and the adjacent side(h) approaches 90, cos t approaches zero because the adjacent side (h) gets shorter and shorter while the opposite side (b = base of triangle) gets longer and longer

sin 90 = 1

cos 90 = 0


for dA/dt = 0

2 * t = 90

t = 45 degrees


substituting the values for C = 2r and t = 45

h = C * cos t

h = 2r * cos 45

h = 2r * sqrt(2)


b = C * sin t

b = 2r * sin 45

b = 2r sqrt(2)


The area is maximum when t = 45 degrees

and

b = h = 2r sqrt(2)

which is an ISOSCELES right triangle.