Sure, let's solve this problem step-by-step:
1. Convert the given dimensions of the smaller box to feet.
The dimensions of the smaller box are:
- Length: 2 ft 5 in
- Width: 2 ft 5 in
- Height: 1 ft 6 in
Let's convert these dimensions from feet and inches to just feet:
- There are 12 inches in a foot.
So, for the length:
[tex]\[
2 \text{ feet } + \frac{5 \text{ inches}}{12} \text{ feet/inch} = 2 + \frac{5}{12} \text{ feet} = 2.4167 \text{ feet}
\][/tex]
Similarly, for the width:
[tex]\[
2 \text{ feet } + \frac{5 \text{ inches}}{12} \text{ feet/inch} = 2 + \frac{5}{12} \text{ feet} = 2.4167 \text{ feet}
\][/tex]
And for the height:
[tex]\[
1 \text{ foot} + \frac{6 \text{ inches}}{12} \text{ feet/inch} = 1 + \frac{6}{12} \text{ feet} = 1.5 \text{ feet}
\][/tex]
2. Calculate the volume of the smaller box.
Using the dimensions now in feet, the volume [tex]\( V \)[/tex] of a rectangular box is given by:
[tex]\[
V = \text{length} \times \text{width} \times \text{height}
\][/tex]
So the volume of the smaller box is:
[tex]\[
V_{smaller} = 2.4167 \text{ ft} \times 2.4167 \text{ ft} \times 1.5 \text{ ft}
= 8.7604 \text{ cubic feet}
\][/tex]
3. Use the given volume ratio to find the volume of the larger box.
The ratio of the volumes of the smaller box to the larger box is 4:9. This means:
[tex]\[
\frac{V_{smaller}}{V_{larger}} = \frac{4}{9}
\][/tex]
To find the volume of the larger box ([tex]\( V_{larger} \)[/tex]), we can set up the following equation:
[tex]\[
V_{larger} = V_{smaller} \times \left( \frac{9}{4} \right)
\][/tex]
Substitute the volume of the smaller box:
[tex]\[
V_{larger} = 8.7604 \times \left( \frac{9}{4} \right) \approx 19.7109 \text{ cubic feet}
\][/tex]
So, the volume of the larger box is approximately:
[tex]\[
19.7109 \text{ cubic feet}
\][/tex]
