Expert: Alon Mandes - 8/4/2009

Question
Water is being collected from a block of ice with a square base. The water is produced because the ice is melting in such a way that each edge of the base of the block is decreasing at 2 inches per hour while the height of the block is decreasing at 3 inches per hour. What is the rate of flow of water into the collecting pan when the base has an edge length of 20 inches and the height of the block is 15 inches? (Make the simplifying assumption that water and ice have the same density.)

How can this problem be solved with out the dimensions of the block before it starts to melt? Or are the 20inx20inx15in dimensions the beginning dimensions and the ice block melts from there?

The answer I came to for the flow of water into the pan was 12in^3/hr. (2in x 2in x 3in = 12in^3/hr) Is this correct?

Also, what does the information in the parentheses have to do with this problem? Or is it there to just throw the student off?

Answer
1st of all , let's translate the given facts into mathematical data :
If x is the edge of the square base, & h is the height then :
1. dx/dt=x'(t)=-2
2. dh/dt=h'(t)=-3
Now, we know that the melting ice is transformed into water. The information in the
parentheses tells us that the melting ice is converting into water in the same rate. Which is
also means that: The rate of change in the melting ice process EQUALS the rate of change in
flowing water.  
Now let's find out how the volume of the ice block is changing :
V(t)=x²(t)h(t) . Let's differentiate it to find rate of change :
dV/dt=V'(t)=2x(t)x'(t)h(t)+x²(t)h'(t) .
The amount of flowing water is the same as V(t). Thus all we need to do now is to calculate
V'(t) when the edge length is 20 ("that means x(t)=20 ") & the height is 15 .
V'=2*20*(-2)*15+20²(-3)
V=-80-1200=-1280. Therefore the change in water flowing is +1280 In³/h .

Alon.