calculus help?

Question

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 10 km/h. Another boat has been heading due east at 20 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?

A farmer with 780 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens.

Area = ft^2

Answer 1

1. let + x be East and - y be South.
At 2PM boat #2 is 20 km East with a velocity of - 20 kph.
Distance between the boats is
d^2 = x^2 + y^2
2ddd = 2xdx + 2ydy = 0 for minimum distance.
xdx/dt = - y dy/dt
-20x = 10y
y = - 2x
x = 20 - 20t
y = -10t
x = 5t = 20 - 20t
25t = 20
t = 0.8 hr = 48 min.

check:
@ t = 0.8 hr, x = 4 km, y = - 8 km
d ≈ 8.94427
@ t = 0.75 hr, x = 5 km, y = - 7.5 km
d ≈ 9.013878
@ t = 0.85 hr, x = 3 km, y = - 8.5 km
d ≈ 9.013878

2. Let W = width and L = length of the area
A = WL
5W + 2L = 780 ft.
5W + 2A/W = 780 ft.
2A = 780W - 5W^2
2dA/dW = 780 - 10W = 0 for max area.
W = 78 ft.
2L = 780 - 390 = 390
L = 195 ft.
A(max) = 78*195 = 15,210 ft^2

Check:
77.9(780 - 5*77.9)/2 = 15209.975
78.1(780 - 5*78.1)/2 = 15209.975