Finding the total surface area of cuboid model using volume I am stuck with a math question about application of differential calculus from my university math class. 


  You are to design a cuboid model with a square base that has a volume of $12m^3$. In order to have a minimum total surface area for the cuboid, what are the values for the height of the cuboid and the length of a side in the square base?


So far, this is what I have done:

$$\text{Volume = }L.B.H = B^2.H$$
$$12 = B^2.H$$
$$H = \frac{12}{B^2}$$
$$\text{S.A. = } 2.B^2 + 4.B.H$$
$$= 2.B^2 + 4.B.(\frac{12}{B^2})$$
$$= 2.B^2 + \frac{48}{B}$$

I am stuck after this step. How should I go about to solve this question? I feel that there is not enough information to complete this question.
In order to find the minimum value of an expression, we take its derivative and equate it to zero. So continuing your solution, and taking the derivative of surface area, we get,

$$\frac{dB}{dy}=4B-\frac{48}{B^2}$$

Equating it to zero, we get

$$B=12^{\frac{1}{3}}$$

Substituting this in:

$$H = \frac{12}{B^2}$$

we get,

$$H=12^{\frac{1}{3}}$$

And $H=L$, so,

$$L=12^{\frac{1}{3}}$$

Finally $L=12^{\frac{1}{3}}, B=12^{\frac{1}{3}}$ and $H=12^{\frac{1}{3}}$
In order to find the minimum value of an expression, we take its derivative and equate it to zero. So continuing your solution, and taking the derivative of surface area, we get,

$$\frac{dB}{dy}=4B-\frac{48}{B^2}$$

Equating it to zero, we get

$$B=12^{\frac{1}{3}}$$

Substituting this in:

$$H = \frac{12}{B^2}$$

we get,

$$H=12^{\frac{1}{3}}$$

And $H=L$, so,

$$L=12^{\frac{1}{3}}$$

Finally $L=12^{\frac{1}{3}}, B=12^{\frac{1}{3}}$ and $H=12^{\frac{1}{3}}$