#
Problem
of the Week

Problem
C and Solution

Ice
Box

## Problem

A metal box in the form of a rectangular prism has an \(18\) cm by \(22\) cm base and a height of \(77\) cm. The box is to be filled with
water, which will then be frozen. When water freezes it expands by
approximately \(10\%\). Determine the
maximum depth to which the box can be filled with water so that when the
water freezes, the ice does not go above the top of the container.

## Solution

**Solution 1**

To determine the volume of a rectangular prism, we multiply its
length, width, and height together. So, the maximum volume of the metal
box is \[18\times 22\times 77=30\,492 \text{
cm}^3\] Let the original depth of water in the metal box be \(h\) cm.

The water volume before freezing is \(18\times 22\times h=(396\times h)\text{
cm}^3\). After the water freezes, the volume increases by \(10\%\) to \(110\%\) of its current volume. That is,
after freezing the volume is \[110\% \text{
of } 396 \times h =1.1\times 396\times h=(435.6\times h)\text{
cm}^3\]

But the volume after freezing is the maximum volume, \(30\,492\text{ cm}^3\). Therefore, \(435.6\times h=30\,492\) and it follows that
\(h=30\,492\div 435.6=70\) cm.

Therefore, the maximum depth to which the box can be filled is \(70\) cm.

**Solution 2**

In this solution we note that the length and width remain the same in
the volume calculations before and after the water freezes. We need only
concern ourselves with the change in the depth of the water.

Let the original depth of water in the container be \(h\) cm.

After freezing, the depth increases by \(10\%\) to \(110\%\) of its depth before freezing. So,
after freezing the depth will be \(110\%
\text{ of } h=1.1\times h=77 \text{ cm}\), the maximum height of
the container. Then \(h=77\div 1.1=70\)
cm.

Therefore, the maximum depth to which the box can be filled is \(70\) cm.