Mathematico | Mathematico

A cuboid has sides of length $2$ , $(x + 1)$ and $(x + 2)$ . The volume of the cuboid is $13$ .

The surface area of the cuboid can be given in simplest form as $a + b \sqrt{c}$ , with $a, b, c \in N$ .

Hint

After expressing the information as a quadratic equation, it helps to divide by $2$ so that the coefficient of $x^{2}$ becomes $1$ .

Solution

By the formula for the volume of a cuboid,

\begin{aligned} 2 (x + 1) (x + 2) & = 13 \\ x^{2} + 3 x + 2 & = \frac{13}{2} \\ {(x + \frac{3}{2})}^{2} - \frac{9}{4} + 2 & = \frac{13}{2} \\ {(x + \frac{3}{2})}^{2} & = \frac{27}{4} \\ x & = - \frac{3}{2} \pm \frac{3 \sqrt{3}}{2} \\ = \frac{- 3 \pm 3 \sqrt{3}}{2} \end{aligned}

It is not possible that $x = \frac{- 3 - 3 \sqrt{3}}{2}$ because then $x + 1$ , a side of the cuboid, would be negative. So $x = \frac{- 3 + 3 \sqrt{3}}{2}$ is the solution we need.

Therefore, the sides have length

\begin{aligned} x + 1 & = \frac{3 \sqrt{3} - 1}{2} \\ x + 2 & = \frac{3 \sqrt{3} + 1}{2} \end{aligned}

and the surface area is

\begin{aligned} 2 (2 \times \frac{3 \sqrt{3} - 1}{2} + \frac{3 \sqrt{3} - 1}{2} \times \frac{3 \sqrt{3} + 1}{2} + \frac{3 \sqrt{3} + 1}{2} + 2) \\ = & 13 + 12 \sqrt{3} \end{aligned}