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.
