I want to minimize the surface area of a rectangular prism, with a constant volume. The dimensions of the prism is $25 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm}$. If it is "flattened out" [with the top face not included], then the dimensions of the "flattened out" prism is $41\text{ cm} \times 24 \text{ cm}$.
Then, $l = 41 - 2c$, $w = 24 - 2c$, and $h = c$. When I use the formula to get the surface area and differentiate with respect to $c$, all $c$ terms vanish. What did I do wrong?
The process that I need to do is differentiation. I think it will be a minimum if $l = w = h$ using AM-GM, but this will not be counted.
Edit: I noticed that the only values of $c$ satisfying $c(41 - 2c)(24 - 2c) = 1600$ are $\left\{8, \frac{49 - \sqrt{1601}}{2}, \frac{49 + \sqrt{1601}}{2}\right\}$. Since $0 \leqslant c \leqslant 12$, then the only values are the first two in the list. When I substituted the second value to the formula for the surface area, it is approximately $1675.906\,\mathrm{cm}^{2}$. Then, the minimum surface area must be $728\,\mathrm{cm}^{2}$. Is this correct?
Edit [March 03, 2021]: $c$, $24 - 2c$, and $41 - 2c$ is now ignored. Dimensions are now free to take any form satisfying the constraints, accepted answer.
If the question is to find dimensions of a rectangular box, with constant volume of $1600$, that minimizes the surface area of the box. The box is closed on top.
Surface area $S = 2l w + 2 l h + 2 w h$
$V = 1600 = lwh \implies h = \frac{1600}{wl}$
So, $S = 2lw + \frac{3200}{w} + \frac{3200}{l}$.
Now take partial derivative wrt $w$ and $l$ and equate to zero to find $w, l, h$ that minimizes surface area.
$\displaystyle S'_w = 2l - \frac{3200}{w^2} = 0 \implies l = \frac{1600}{w^2}$
$\displaystyle S'_l = 2w - \frac{3200}{l^2} = 0 \implies w = \frac{1600}{l^2}$
Solving both, $\displaystyle l = w = 4 \sqrt[3]{25}$. Then use this to find value of $h$ and you should get a cube.