Find minimum distance?

Question

Find Minimum Distance..?
At noon, ship A is 100 miles due east of ship B. Ship A sailing west at 12 miles/hr, and ship B is sailing south at 10 miles/hr. At what time will the ships be nearest to each other, and what will this distance be?

2) Find the length of the longest pipe that can be carried level around a right-angle corner if the two intersecting corridors are of widths 4 feet and 6 feet. (Dont use trigonometry)

Answer 1

These are optimization problems. The most difficult part (I think) is identifying the equation that describes what you want to maximize or minimize.

Question 1:

A starts out to the right of B 100 miles away.

Define the following:
x = the distance A travels eastward (towards the left)
100 - x = the distance between the new location of A and the original location of B
y = the distance B travels southward
z = the distance between the new location of A and new location of B

Note that new A, new B and old B forms a right triangle with legs of length 100 - x and y and a hypotenuse of length z.

Then (100 - x)^2 + y^2 = z^2 by Pythagorean's Theorem.

Note that z^2 is what you want to minimize (i.e. "minimizing" distance is the same thing as determining the "nearest" possible distance) and that minimizing z^2 is the same thing as minimizing z itself.

The only problem is that z^2 is in terms of two variables: x and y. Taking derivatives with respect both variables would be extremely complicated. So let's reduce this equation down to be only in terms of x.

To get y in terms of x, we consider the classic RTD formula:
rate x time = distance

Suppose t = the number of hours after noon that it takes for A and B to be "closest" to each other. Then:
12*t = x => t = x/12
10*t = y => t = y/10

Thus, x/12 = y/10 => y = (10/12)x = (5/6)x

Substituting this into our z^2 formula:

z^2 = (100 - x)^2 + ((5/6)x)^2

To find the minimum, we take the derivative of z^2 with respect to x and find critical points.

(z^2)' = -2(100 - x) + (25/18)x = -200 + (61/18)x

Critical points occur when (z^2)' = 0 or (z^2)' DNE.

In this case, -200 + (61/18)x = 0 => x = 3600/61

x < 3600/61 => (z^2)' < 0 => z^2 is decreasing
x > 3600/61 => (z^2)' > 0 => z^2 is increasing

A function that decreases and then increases must achieve a minimum at the critical point (by the first derivative test).

So x = 3600/61 will give us our needed minimum distance.

But at what time does this occur?
t = x/12 = 3600/(61*12) = 4.92
i.e. it occurs about 4.92 hours after noon = 4:55pm

How far apart are they at 4:55pm?

z^2
= (100 - x)^2 + ((5/6)x)^2
= (100 - 3600/61)^2 + ((5/6)*(3600/61))^2
= 4098.36

So z = 64.02 miles

Answer 2

Let

a = east/west distance between the two ships
b = north/south distance between the two ships
c = distance between the two ships
t = time elapsed

da/dt = -12 miles/hour
db/dt = 10 miles/hour

a = 100 - 12t
b = 10t

c² = a² + b² = (100 - 12t)² + (10t)²
= 10,000 - 2400t + 144t² + 100t²
= 244t² - 2400t + 10,000

2c(dc/dt) = 488t - 2400
(dc/dt) = (488t - 2400)/2c
(dc/dt) = (244t - 1200)/c = 0
244t = 1200
t = 1200/244 = 300/61 = Noon + 300/61 hours ≈ 4:55:05 pm

a = 100 - 12t = 100 - 12(300/61) = (6100 - 3600)/61 = 2500/61
b = 10t = 10(300/61) = 3000/61

c = √(a² + b²) = √{(2500/61)² + (3000/61)²}
c = (100/61)√(25² + 30²) = (100/61)√(625 + 900)
c = (100/61)√1525 = (500/61)√61 = 500/√61 = 64.01844 miles

The ships will make their closest approach of 64.01844 miles apart at 4:55:05 pm.