Remember that the volume of a rectangular solid is length × width × height. Substitute your expressions for length and height (in terms of $w$) into this formula and set it equal to 216.

After you write the volume equation, rearrange it into a standard form (everything on one side equals 0). Look for an integer value of $w$ that makes the equation true and also keeps all dimensions positive.

In Desmos, type the equation 2x^2(x-3) = 216. Desmos will automatically treat this as an equation to graph.

If needed, instead enter y = 2x^2(x-3) - 216 so that the solutions correspond to the x-intercepts of the graph (where $y=0$).

Look at the graph and find the x-value where the curve crosses the x-axis and where both $x$ and $x-3$ are positive. That x-value is the width of the solid.

Let the width of the rectangular solid be $w$ inches.

Then, using the information in the problem:

• The length is twice the width, so the length is $2w$.
• The height is 3 inches less than the width, so the height is $w-3$.

The volume $V$ of a rectangular solid is given by

We are told the volume is 216 cubic inches, so

This simplifies to

Start solving for $w$ by dividing both sides by 2:

Rewrite this as

Distribute $w^2$:

Now we need to find a value of $w$ that makes this equation true.

Because the volume is a nice integer, it is reasonable to test integer values of $w$ that are factors of 108.

Notice that $216$ is a perfect cube, and $\sqrt[3]{216} = 6$, so try $w=6$ in the equation $w^3 - 3w^2 - 108 = 0$:

• $w^3 = 6^3 = 216$
• $3w^2 = 3 \cdot 6^2 = 3 \cdot 36 = 108$

Then

so $w=6$ satisfies the equation. Since width must be positive and also make the height $w-3$ positive, this value is valid. Therefore, the width of the solid is $6$ inches.