The surface area of an open box with a square base in 36 units. It's maximum volume (in cubic units) is:

The correct answer is 12√3.

We are given an open box (no lid) with a square base, whose total surface area is 36 square units. We need to find the dimensions that maximize the volume.

Step 1: Set Up Variables

Let the side length of the square base be x, and the height of the box be h.

Step 2: Write the Surface Area Constraint

An open box has one square base and four rectangular sides (no top). So:

$x^2+4xh=36$

Solving for h:

$h=\frac{36-x^2}{4x}$

Step 3: Express Volume as a Function of x

The volume of the box is:

$V=x^2·h=x^2·\frac{36-x^2}{4x}=\frac{x(36-x^2)}{4}$

So:

$V\left(x\right)=\frac{36x-x^3}{4}$

Step 4: Differentiate and Find Critical Points

Differentiate V with respect to x and set equal to zero:

$\frac{dV}{dx}=\frac{36-3x^2}{4}=0$

$36-3x^2=0\Rightarrow x^2=12\Rightarrow x=2\sqrt{3}$

Step 5: Verify it is a Maximum

$\frac{d^{2}V}{d{x}^2}=\frac{-6x}{4}$
At $x=2\sqrt{3}$, this is negative, confirming a maximum.

Step 6: Find the Optimal Height h

$h=\frac{36-x^2}{4x}=\frac{36-12}{4·2\sqrt{3}}=\frac{24}{8\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$

Step 7: Calculate the Maximum Volume

$V=x^2·h=12·\sqrt{3}=12\sqrt{3}$

Therefore, the maximum volume of the open box is 12√3 cubic units. This is achieved when the square base has side length $2\sqrt{3}$ and height $\sqrt{3}$, which interestingly means the height is exactly half the base side length — a classic hallmark of optimized open-box problems.