As the other answer suggests, if we don't place any restriction on
what values the sides of a cuboid can take, there are way too many possibilities.
We will assume the sides of the cuboid are all integers.
Given a cuboid of dimension $a \times b \times c$ where $a,b,c$ are integers,
we have
$$\begin{align}
\verb/Area/ &= 2(ab+bc+ca)\\
\verb/Vol/ &= abc\\
\end{align}$$
WOLOG, we will assume $a \le b \le c$ and rewrite the formula for area as
$$(a+b)(a+c) = (ab+bc+ca)+a^2 = \frac12\verb/Area/+a^2 = 120+a^2$$
Since $$\verb/Area/ = 240 \implies 120 = (ab+bc+ca) \ge 3a^2 \implies
1 \le a \le 6$$
We have only $6$ cases to analysis.
$$\begin{array}{clll}
a = 1 &\implies 120+a^2 = 121 = 11\times 11 &\leadsto& (a,b,c,\verb/Vol/) = (1,10,10,100)\\
a = 2 &\implies 120+a^2 = 124 = 4 \times 31 &\leadsto& (a,b,c,\verb/Vol/) = (2,2, 29,116)\\
a = 3 &\implies 120+a^2 = 129 = 3 \times 43 &\leadsto& \verb/nothing/\\
a = 4 &\implies 120+a^2 = 136 = 2^3 \times 17 &\leadsto& (a,b,c,\verb/Vol/) = (4,4,13,208)\\
a = 5 &\implies 120+a^2 = 145 = 5 \times 29 &\leadsto& \verb/nothing/\\
a = 6 &\implies 120+a^2 = 156 = 2^2 \times 3 \times 13 &\leadsto&
(a,b,c,\verb/Vol/) = (6,6,7,252)
\end{array}$$
To summarize, if the sides of the cuboids are all integers, there are 4 possible
volumes $100, 116, 208, 252$.
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.
Best Answer
While there are a few ways to go about solving this problem, there are surely a few mistakes in the work shown here, but they can be quickly solved: We can assume that because it is a triangular prism on top of a rectangular prism that the roof will not be considered in the "outside" surface area of the Greenhouse (I'm following the idea that we will later use this surface area relative to the sunlight coming in through this amount of surface area, therefore only considering the outside surface area). With this same logic, the surface area of the floor should not be considered either. Here is what we must solve for in this problem now:
Please note that you did not specify in the original problem presentation the orientation of the triangular prism on top of the rectangular prism, so instead of assuming the triangles would be placed above the short sides of the green house, they could be above the longer sides, instead. (If you run into further issues, assumptions about dimensions may be withholding the correct answer)
Moving on from this, let's solve for the four parts of this figure:
A - Long sides perpendicular to ground
Length: 96', Height: 6 1/3', Area = 96*(19/3) = 32*19 = 608 * 2 (two sides) = 1216 ft^2
B - Short sides perpendicular to ground
Length: 23', Height, 6 1/3', Area = 23*(19/3) = 437/3 = 145.667 * 2 (two sides) = 291.333 ft^2
C - Triangle sides of the triangular prism (assuming above the short side of the rectangle, as proposed in the OP)
Please Note: In the OP, you assumed that the triangle was a right triangle, then using the Gougu (Pythagorean) Theorem, not the Law of Cosines, to find the length of the legs of the isosceles triangle. This triangle is actually not a right triangle:
Height (altitude from vertex adjoined to equal side lengths) = 12' - 6 1/3' = 5 2/3' . This is perpendicular to the long side of the triangle, being 23' under the OP's assumption; therefore, the area of the triangle is just (bh)/2 (not bh as shown in the OP) . Calculating this area is simply just: 23' x 5 2/3' = (23 x 17)/(3*2) * 2 (two sides) = 130.333 ft^2
D - Slanted ceiling surface rectangles
The length of these rectangles is 96' (as assumed in the OP), and we can find the height (triangle's shorter leg length) using the Gougu Theorem:
The base of the isosceles triangle is 23', and the height is 5 2/3'. Divide the base by 2 to find the leg of the desired right triangle (with the slanted ceiling's side length being the hypotenuse of the aforementioned right triangle). There is now a right triangle with base of 11 1/2' and height of 5 2/3'. Using the Gougu theorem, we get 11.5^2 + 5.667^2 = h^2, h = sqrt(132.25 + 32.111) = 12.8203 ft.
Now the length of the rectangle, 96', times the height, 12.8203', is 1230.7526 * 2 (two sides) = 2461.5 ft^2
Final Answer! - Summing all of these parts together, we get A + B + C + D = 4099
Thanks for the fun problem, and I hope that this helps! Go follow my Youtube Channel as I will be updating it in the future.