Hello nthabiseng p,
Let's first verify the given hint is true. For this I suggest using Lagrange multipliers. I will let the non-negative dimensions of the cuboid be $x,y,z$.
We have the objective function:
$$V(x,y,z)=xyz$$
subject to the constraint:
$$g(x,y,z)=2xy+2xz+2yz-S=0$$ where $S$ is the surface area to which we are restricted.
Thus, we find the system:
$$yz=(2y+2z)\lambda$$
$$xz=(2x+2z)\lambda$$
$$xy=(2x+2y)\lambda$$
This system implies that for non-zero dimension values, we must have:
$$x=y=z$$
So we know the given hint is true. Substituting into the constraint, we find:
$$6x^2=S\,\therefore\,x=\sqrt{\frac{S}{6}}$$
Thus, we should find the dimensions of the cuboid having a given surface area $S$ to be:
$$x=y=z=\sqrt{\frac{S}{6}}$$
Now, assuming we are to use single-variable calculus instead, along with the given hint ($z=x$), we could write:
$$V=x^2y$$
$$S=2x^2+4xy$$
Solving the second equation for $y$, we find:
$$y=\frac{S-2x^2}{4x}$$
Substituting for $y$ into the first equation, there results:
$$V(x)=x^2\left(\frac{S-2x^2}{4x} \right)=\frac{1}{4}\left(Sx-2x^3 \right)$$
Now, differentiating and equating to zero to find the critical point, we get:
$$V'(x)=\frac{1}{4}\left(S-6x^2 \right)=0$$
Since we require $$0\le x$$ we get the critical value:
$$x=\sqrt{\frac{S}{6}}$$
Now, using the second-derivative test, we find:
$$V''(x)=-3x$$
We see the function is concave down for any positive value of $x$, thus we know our critical value is at a maximum, which is global.
And so, using the value for $y(x)$, we find:
$$y=\frac{S-\frac{S}{3}}{4\sqrt{\frac{S}{6}}}=\sqrt{\frac{S}{6}}$$
Hence, we find that:
$$x=y=z=\sqrt{\frac{S}{6}}$$
Using the value $$S=6200\text{ cm}^2$$ we then find:
$$x=y=z=\frac{10\sqrt{93}}{3}\,\text{cm}$$.