Mathematico | Mathematico

An open-topped cuboid is constructed. The base has width $b$ and depth $2 b$ and the height is $h .$ The length of the main diagonal of the cuboid is known to be $5 cm$ in length.

Given that the cuboid is constructed so that the volume is as large as possible, find the surface area of the exterior of the cuboid.

First, use Pythagoras in three dimensions to show that

b^{2} = 5 - \frac{h^{2}}{5}

Pythagoras in three dimensions works like this:

ℓ^{2} = x^{2} + y^{2} + z^{2}

Substitute

b^{2} = 5 - \frac{h^{2}}{5}

into the volume

V = 2 b \times b \times h = 2 b^{2} h

and then solve

\frac{d V}{d h} = 0

to find the $h$ for which the maximum volume occurs.

You can substitute the value of $h$ we found into

b^{2} = 5 - \frac{h^{2}}{5}

to find the corresponding value of $b$ and then put them both into the formula for the area.

Remember, the cuboid has no top, so the area is composed of $5$ rectangles rather than $6.$

Give your answer to $3$ significant figures.