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35 yachts
Try this math question...it's good for ur brain?
1. A passenger jet is travelling due north at 400km/h, while a cargo plane, 2km to the northwest, is travelling east at 300km/h. The altitude of the cargo plane is 1000 m lower than the altitude of the passenger jet. What will be the minimum separation between the two aircraft.
Ans: 1.039 km
2. Lisa, who is standing on the deck of a yacht that is travelling due west at 6km/h, see a sailboat sailing southwest at 4km/h, 3 km northwest of the yacht. How close to each other do these boats get?
Ans: 0.171 km
3. Bill owns an oil well located 400m from a road. Bill wansts to connect the well to a storage tank 1200 m down the road from the well. It costs $35/m to lay pipe along the road and $50/m to lay it elsewhere. How should the pipeline be laid to minimize the total cost?
Ans: lay 807.9m along the road 560m
some of the answers might be wrong =s
Scranderberry you are a genius!
Now just wait for another 10 points! You surely deserve it!
1.
I don't know if this is the conventional way to solve this problem, but I looked at the 3d coordinates of the planes
If the centre of the compass at time 0 sits at the centre of the xy axis, then the passenger jet is at (0, 400t, 0)
Where t is time, and assuming the passenger jets height is 0.
Now the cargo plane position is 2km to the northwest, if you make a triangle, that √2km west and √2km north
So the position of the cargo plane is:
(-√2 + 300t, √2, -1)
It moves 300km per hour is the positive x direction from x = -√2, it's y coordinate doesn't change and stays at √2, and its z coordinate is 1 km below 0)
Now the distance between the planes is given by the following formula
(-√2 + 300t, √2, -1)
(0, 400t, 0)
d = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)²
= √((-√2 + 300t - 0)² + (√2 - 400t)² + (-1 - 0)²
=√( 2 - 600t√2 + 90000t² + 2 - 800t√2 + 160000t² + 1)
= √(250000t² - 1400t√2 +5)
d ' = (500000t - 1400√2) / (2√(250000t² - 1400t√2 +5)) = 0
500000t = 1400√2
t = 0.0039598hours (after only 14.25seconds)
The distance is:
= √(250000t² - 1400t√2 +5)
= 1.039km (3dp)
2.
Ok, I think my original solution must have had errors int he squaring and square roots, so I've avoided those this time.
Lisa's yacht = (-6t, 0)
Sailboat = (-3/√2 - 2t√2, 3/√2 - 2t√2)
To explain this - the sailboat starts at (-3 / √2, 3 / √2) - thats 3 km northwest of the origin, and every hour it travels 4km/hr southwest - which for every hour is 2t√2 left along the x axis, and 2t√2 down on the y axis)
Now rather than minimising d I will minimise d² (same answer)
d² = (x2 - x1)² + (y2 - y1)²
= (-6t - (-3/√2 - 2t√2))² + (0 - (3/√2 - 2t√2))²
= (3/√2 + 2t√2 - 6t)² + (2t√2 - 3/√2)²
d² ' = 2(2√2 - 6)(3/√2 + 2t√2 - 6t) + 2(2√2)(2t√2 - 3/√2) = 0
divide by 2 and expand
(2√2 - 6)(3/√2 + 2t√2 - 6t) + (2√2)(2t√2 - 3/√2) = 0
6 + 8t - 12√2t - 18/√2 - 12√2t + 36t + 8t - 6 = 0
(8 - 12√2 - 12√2 + 36 + 8)t = 18/√2
t = (18/√2) / (8 - 12√2 - 12√2 + 36 + 
= 0.7hours (1 dp)
d = √((3/√2 + 2t√2 - 6t)² + (2t√2 - 3/√2)²)
= 0.171km (3dp)
3.
ok, let x = distance from the road next to the well, to where the pipe meets the road
Cost = 35(1200 - x) + 50√(x² + 400²)
C' = -35 + 50*2x / 2√(x² + 400²) = 0
50x / √(x² + 400²) = 35
10x = 7√(x² + 400²)
100x² = 49(x² + 400²)
51x² = 49*400²
x² = 49*400² / 51
x = 392.08m (2dp)
so lay 1200 - x = 807.9m along the road and √(x² + 400²) = 560m elsewhere (nearest m)
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{ 2 comments }
for the second question use x=v(i)t+(1/2)at^2
you are looking for x and you have the other info
v(i)=0
t=86400s (that much seconds in a day)
for the last question use
v(f)^2=v(i)^2+2ax
solve for v(f)
Actually the boat being offered as this prize is a J105 which is, indeed, a high performance sailing craft compared to 90% of 35' yachts.
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