# Simultaneous equation (?) word problem

• Nov 1st 2010, 04:22 AM
marmarzet
Simultaneous equation (?) word problem
I'm struggling with this...

"Three types of ship are used to transport goods, type P, Q and R. The ships are the same shape and design, but have different sizes. Their cargo capacity depends on the hold volume. The number of crew required is proportional to the surface area of the deck. Types Q and R taken together have the same length as 2 type Ps. The crew of a type Q is just sufficient to provide crew for two type Ps and a type R.

A fully loaded type Q wants to transfer all of its cargo to smaller Ps and Rs, while minimising the nymber of crew needed for the resultant fleet. How many type Ps and Rs are needed?"

It also gives a hint, saying that for obhects of any shape the surface area is proportional to the square of the object's size and the volume proportional to the cube of its size.

I'd really appreciate any help.

Thank you!
• Nov 1st 2010, 04:51 PM
emakarov
Interesting problem. By abuse of notation, let the lengths of the ships P, Q, and R also by P, Q, and R, respectively. I chose variables to denote lengths rather than areas or volumes to avoid square or cubic roots. Then the problem says that Q + R = 2P, i.e., P is the arithmetic mean of Q and R. Since we are also told that Q is the bigger of the three ships, we have R < P < Q.

Since the number of crew is proportional to the deck area, which itself is proportional to the square of length, the crews of P, Q, and R are $cP^2$, $cQ^2$ and $cR^2$, respectively, for some unknown coefficient $c$. The problem tells us that $cQ^2 = 2cP^2+cR^2$, or $Q^2 = 2P^2+R^2$.

Since the volume grows faster with linear size than the number of crew, it is better to transfer cargo from Q to as many intermediate-sized P's as possible and use the smallest R's only for the remainder. So let us find Q / P. From the two equations above, Q / P = 3 / 2 in my calculations (needs double-checking). Similarly, Q / R = 3 (same remark). Therefore, the cargo of Q can be transferred to 1 P and 1 R.
• Nov 2nd 2010, 10:02 AM
marmarzet
Thank you!

I get it all up to the point where you say Q/P=3/2 and Q/R=3 therefore P=1 and R=1. Why does that follow?

Thanks
• Nov 2nd 2010, 10:21 AM
Unknown008
When you solve the two equations eliminating R, you get:

$Q^2 = 2P^2 + (2P - Q)^2$

Simplifying this gives:

$\dfrac{Q}{P} = \dfrac32$

Similarly, eliminating P, we get the equation: $Q^2 = 2\left(\dfrac{Q+R}{2}\right)^2 + R^2$

$\dfrac{Q}{R} = 3$

Put this on a scale, and we get:

$R : P:Q = \dfrac{Q}{3} : \dfrac{2Q}{3} : Q$

From this, we can deduce that 1 P and 1 R will be enough.
• Nov 2nd 2010, 10:42 AM
marmarzet
Ah, I see. Thank you both!
(Happy)
• Nov 2nd 2010, 10:48 AM
Unknown008
You're welcome (Happy)