1. Form the volume function

After cutting squares of side $x$ from each corner of the sheet $a \times b$ and folding, the box has:

• length $= a-2x$
• breadth $= b-2x$
• height $= x$

So the volume is

$V(x)=x(a-2x)(b-2x).$

2. Expand the expression

$V(x)=x\big(ab-2ax-2bx+4x^2\big)$

$V(x)=abx-2(a+b)x^2+4x^3.$

3. Differentiate to maximize volume

$V'(x)=ab-4(a+b)x+12x^2.$

For maximum volume,

$V'(x)=0$

$12x^2-4(a+b)x+ab=0.$

Divide by $4$:

$3x^2-(a+b)x+\frac{ab}{4}=0.$

Using the quadratic formula on

$12x^2-4(a+b)x+ab=0,$

we get

$x=\frac{4(a+b)\pm \sqrt{16(a+b)^2-48ab}}{24}.$

Simplify the discriminant:

$16(a+b)^2-48ab=16\big((a+b)^2-3ab\big)$

$=16(a^2+2ab+b^2-3ab)$

$=16(a^2-ab+b^2).$

Hence,

$x=\frac{4(a+b)\pm 4\sqrt{a^2+b^2-ab}}{24}$

$x=\frac{a+b\pm \sqrt{a^2+b^2-ab}}{6}.$

4. Choose the physically valid value

Since $x$ must be small enough so that $a-2x>0$ and $b-2x>0$, we take the smaller root:

$x=\frac{a+b-\sqrt{a^2+b^2-ab}}{6}.$

5. Check that it gives maximum

$V''(x)=-4(a+b)+24x.$

At the smaller critical point, $V''(x)<0$, so the volume is maximum there.

6. Compare with options

This matches:

$\boxed{\frac{a+b-\sqrt{a^2+b^2-ab}}{6}}$

So the correct option is C.