Calculus/Optimization questions? * BEST ANSWER POINTS

1. Three sides of a trapezoid are 10 cm each. What is the length of the fourth side such that the area will be a maximum?

2. A rectangular field is to have an area of 100m^2. It will be enclosed by fencing. The north/south sides will cost $20/m and the east/west sides will cost $5/m. What are the dimensions to minimize the cost?

FIRST PERSON TO ANSWER CORRECTLY WITH WORK WILL RECIEVE 10 PTS.

A = (1/2) * (b[1] + b[2]) * h
b[1] = 10
b[2] = 10 + 2 * sqrt(10^2 - h^2)

A = (1/2) * (10 + 10 + 2 * sqrt(100 - h^2)) * h
A = (1/2) * (20 + 2 * sqrt(100 - h^2)) * h
A = 10h + h * sqrt(100 - h^2)
dA/dh = 10 + h * (100 - h^2)^(-1/2) * (-2h) + (100 - h^2)^(1/2)
dA/dh = 0
0 = 10 - 2h^2 / sqrt(100 - h^2) + sqrt(100 - h^2)
2h^2 / sqrt(100 - h^2) = 10 + sqrt(100 - h^2)
2h^2 = 10 * sqrt(100 - h^2) + 100 - h^2
3h^2 - 100 = 10 * sqrt(100 - h^2)
9h^4 - 600h^2 + 10000 = 100 * (100 - h^2)
9h^4 - 600h^2 + 10000 = 10000 - 100h^2
9h^4 - 500h^2 = 0
h^2 * (9h^2 - 500) = 0

h = 0

9h^2 - 500 = 0
9h^2 = 500
h^2 = 5 * 100 / 9
h = (10 / 3) * sqrt(5)

b[2] = 10 + 2 * sqrt(100 - h^2)
b[2] = 10 + 2 * sqrt(100 - 500/9)
b[2] = 10 + 2 * sqrt((900 - 500) / 9)
b[2] = 10 + 2 * (1/3) * sqrt(400)
b[2] = 10 + 2 * (1/3) * 20
b[2] = (30 + 40) / 3
b[2] = 70/3



L * W = 100
20 * L + 5 * W = C

20 * (100 / W) + 5 * W = C
2000 / W + 5 * W = C
-2000 / W^2 + 5 = dC/dW
dC/dw = 0
-2000 / W^2 + 5 = 0
2000 / W^2 = 5
400 / W^2 = 1
400 = W^2
20 = W
L = 5

5 * 20 + 20 * 5 = 200