Question

a company wants to sell blocks of cheese in the shape of rectangular prisms, as shown.
block of cheese
each block of cheese must have a volume that is a rational number. which of the following can be the length (l), width (w) and height (h) of a block of cheese?
a. l = √2, w = 3√3, h = 2√6
b. l = √2, w = 3, h = 6
c. l = 2, w = 3√3, h = 2√6
d. l = √2, w = 3√2, h = 6√2

Explanation:

The volume \( V \) of a rectangular prism is given by the formula \( V = l \times w \times h \). We need to check which of the given options results in a rational number (a number that can be expressed as a fraction of two integers) for the volume.

Step 1: Recall the formula for the volume of a rectangular prism

The volume \( V \) of a rectangular prism with length \( l \), width \( w \), and height \( h \) is \( V = l \times w \times h \).

Step 2: Analyze Option A

For Option A: \( l = \sqrt{2} \), \( w = 3\sqrt{3} \), \( h = 2\sqrt{6} \)
\[ \begin{align*} V &= \sqrt{2} \times 3\sqrt{3} \times 2\sqrt{6} \\ &= 3 \times 2 \times \sqrt{2} \times \sqrt{3} \times \sqrt{6} \\ &= 6 \times \sqrt{2 \times 3 \times 6} \\ &= 6 \times \sqrt{36} \\ &= 6 \times 6 \\ &= 36 \end{align*} \]
Wait, let's re - calculate the product of the square roots:
We know that \( \sqrt{a}\times\sqrt{b}=\sqrt{ab} \). So,
\( \sqrt{2}\times\sqrt{3}=\sqrt{6} \), then \( \sqrt{6}\times\sqrt{6} = 6 \)
So, \( V=3\times2\times6=36 \) (rational number). Wait, but let's check other options as well.

Step 3: Analyze Option B

For Option B: \( l=\sqrt{2} \), \( w = 3 \), \( h=6 \)
\[ \begin{align*} V&=\sqrt{2}\times3\times6\\ &=18\sqrt{2} \end{align*} \]
\( \sqrt{2}\approx1.414 \), so \( 18\sqrt{2}\approx25.456 \), which is an irrational number (since \( \sqrt{2} \) is irrational and multiplying a non - zero rational number by an irrational number gives an irrational number).

Step 4: Analyze Option C

For Option C: \( l = 2 \), \( w=3\sqrt{3} \), \( h = 2\sqrt{6} \)
\[ \begin{align*} V&=2\times3\sqrt{3}\times2\sqrt{6}\\ &=12\times\sqrt{3\times6}\\ &=12\times\sqrt{18}\\ &=12\times3\sqrt{2}\\ &=36\sqrt{2} \end{align*} \]
\( \sqrt{2}\approx1.414 \), so \( 36\sqrt{2}\approx50.904 \), which is an irrational number.

Step 5: Analyze Option D

For Option D: \( l=\sqrt{2} \), \( w = 3\sqrt{2} \), \( h=6\sqrt{2} \)
\[ \begin{align*} V&=\sqrt{2}\times3\sqrt{2}\times6\sqrt{2}\\ &=3\times6\times\sqrt{2}\times\sqrt{2}\times\sqrt{2}\\ &=18\times2\times\sqrt{2}\\ &=36\sqrt{2} \end{align*} \]
\( \sqrt{2}\approx1.414 \), so \( 36\sqrt{2}\approx50.904 \), which is an irrational number.

Wait, there was a miscalculation in Step 2. Let's re - do Option A:

\( l=\sqrt{2} \), \( w = 3\sqrt{3} \), \( h=2\sqrt{6} \)

\( V=\sqrt{2}\times3\sqrt{3}\times2\sqrt{6}=3\times2\times\sqrt{2}\times\sqrt{3}\times\sqrt{6} \)

We know that \( \sqrt{2}\times\sqrt{3}=\sqrt{6} \), then \( \sqrt{6}\times\sqrt{6} = 6 \)

So, \( V = 3\times2\times6=36 \) (rational number)

Option A gives a rational volume. Wait, but let's check the calculation again.

\( \sqrt{2}\times3\sqrt{3}\times2\sqrt{6}=3\times2\times\sqrt{2}\times\sqrt{3}\times\sqrt{6}=6\times\sqrt{2\times3\times6}=6\times\sqrt{36}=6\times6 = 36 \) (rational)

Option B: \( V=\sqrt{2}\times3\times6 = 18\sqrt{2}\) (irrational)

Option C: \( V = 2\times3\sqrt{3}\times2\sqrt{6}=12\times\sqrt{18}=12\times3\sqrt{2}=36\sqrt{2}\) (irrational)

Option D: \( V=\sqrt{2}\times3\sqrt{2}\times6\sqrt{2}=3\times6\times\sqrt{2}\times\sqrt{2}\times\sqrt{2}=18\times2\times\sqrt{2}=36\sqrt{2}\) (irrational)

So, Option A gives a rational volume. Wait, but let's check the problem statement again. The problem says "Which of the following can be the length (l), width (w) and height (h) of a block of cheese?" with the volume being rational.

Wait, in Option A, when we calculate the volume:

\( l=\sqrt{2}\), \( w = 3\sqrt{3}\), \( h=2\sqrt{6}\)

\( V=\sqrt{2}\times3\sqrt{3}\times2\sqrt{6}=3\times2\times\sqrt{2}\times\sqrt{3}\times\sqrt{6}=6\times\sqrt{2\times3\times6}=6\times\sqrt{36}=6\times6 = 36 \) (rational)

Option B: \( V=\sqrt{2}\times3\times6 = 18\sqrt{2}\) (irrational)

Option C: \( V=2\times3\sqrt{3}\times2\sqrt{6}=12\times\sqrt{18}=12\times3\sqrt{2}=36\sqrt{2}\) (irrational)

Option D: \( V=\sqrt{2}\times3\sqrt{2}\times6\sqrt{2}=3\times6\times\sqrt{2}\times\sqrt{2}\times\sqrt{2}=18\times2\times\sqrt{2}=36\sqrt{2}\) (irrational)

So the correct option is A.

Wait, but let's re - check the calculation for Option A:

\( \sqrt{2}\times3\sqrt{3}=\ 3\sqrt{6}\)

Then \( 3\sqrt{6}\times2\sqrt{6}=3\times2\times\sqrt{6}\times\sqrt{6}=6\times6 = 36 \)

Yes, that's correct. So the volume for Option A is 36, which is a rational number.

Answer:

A. \( l=\sqrt{2},w = 3\sqrt{3},h = 2\sqrt{6}\)