Let the side of each cut-out square be $x$ cm.

Then after cutting and folding:

• length of box $= 30 - 2x$
• breadth of box $= 30 - 2x$
• height of box $= x$

Form the volume function.

$V(x) = x(30-2x)^2$

We need to maximize this volume.

Differentiate $V(x)$.

$V(x) = x(30-2x)^2$

Using product rule:

$\frac{dV}{dx} = (30-2x)^2 + x \cdot 2(30-2x)(-2)$

$\frac{dV}{dx} = (30-2x)^2 - 4x(30-2x)$

$\frac{dV}{dx} = (30-2x)\big[(30-2x)-4x\big]$

$\frac{dV}{dx} = (30-2x)(30-6x)$

Set derivative equal to zero.

$(30-2x)(30-6x)=0$

So, $x=15 \quad \text{or} \quad x=5$

But $x=15$ makes the base side $30-2x=0$, so volume becomes zero. Hence the valid maximizing value is:

$x=5$

Find the dimensions of the box.

$\text{length} = \text{breadth} = 30-2(5)=20 \text{ cm}$ $\text{height} = 5 \text{ cm}$

Find the surface area of the open box.

Since the box has no top, its surface area is:

$\text{Surface area} = \text{base area} + \text{area of 4 sides}$

Base area: $20 \times 20 = 400$

Four sides: $4(20 \times 5)=400$

Therefore, $\text{Surface area} = 400+400=800 \text{ cm}^2$

Compare with options.

The correct option is: $\boxed{\text{C: } 800}$

Comparison with stored correct answer.

Stored correct answer = C

Our derived answer = C

So they agree.