Expert: Paul Klarreich - 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
Questioner: Evan
Country: United States
Category: Calculus
Private: No
Subject: Calculus Question

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 without 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?

>> Irrelevant.

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?

>> which parentheses?
...............................................................
This looks like your standard related-rates problem. (look them up in the site archives -- click on BROWSE PAST ANSWERS.)

Part 1: The things that change are:

x = the width and length of the block at any time (it's a square)
h = the height of the block at any time.
V = ..  volume ........

Part 2: The rates are:
dx/dt = 2
dh/dt = 3
dV/dt TO BE FOUND.

3: The relation is:

V = x^2h

4: Differentiate:

dV/dt = 2xh dx/dt + x^2 dh/dt

5: Substitute
x = 20
h = 15

dV/dt = 2(20)(15)(2) + (20)^2(15)

dV/dt = 1200 + 600 = 1800