The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.
The surface area of the box described is
A=x^2 +4xh#
We need
A# as a function of
x# alone, so we'll use the fact that
V=x^2h = 32,000# cm^3
which gives us
h = [32,000]/x^2#, so the area becomes:
A=x^2 +4x[[32,000]/x^2] = x^2 +[128,000]/x#
We want to minimize
A#, so
A' = 2x-[128,000]/x^2 = 0# when #[2x^3-128,000]/x^2 = 0#
Which occurs when
x^3 - 64,000 = 0# or
x=40#
The only critical number is
x=40# cm.
The second derivative test verifies that
A# has a minimum at this critical number:
A'' = 2+[256,000]/x^3# which is positive at
x = 40#.
The box should have base 40 cm by 40 cm and height 20 cm.
[use
h = [32,000]/x^2# and
x=40#]
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Solution
The correct option is D
lb+2bh+2hl
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I think you are given the image of a box [with no face at the top], where the sides and bottom have only length and width, but no depth [in an ideal topless rectangular cube.
For example, if we align the a $10 \text{cm } \times 10 \text{cm } \times 10\text{cm }$ ideal box [with no top in the $z = 10$ plane, with one corner at the origin, and the bottom and sides that share that corner, along the x y z axes, the first octant, with the top open, point $[2, 3, 0]$ on the "bottom of the box" is the same point on that's on the bottom of the inside of the box.
So taking the surface area, there is really no "outside" or "inside" area. We exclude, unless given, depth of wood [if the box is wooden] or even cardboard [accordingly].
So the surface area of such an "ideal" box [rectangular cube] is the sum of the area of the bottom side, plus the sum of the areas of each of four sides.
In the example I give above, the surface area is $10 \times 10 \times 5 = 500 \text{cm}^3$.
Note, unless told the thickness of a box's bottom and sides, we take that the "external" side is identical to the "internal side". So count only the area of the five sides, summed.
If, on the other hand, you are given that the open-topped box [say a wooden box], with width/length of the bottom, and height/length of the sides, and the thickness of the wood, or metal, or cardboard [say 2mm], then we'd need to take the surface area of the inside, which would necessarily be less than the external surface area [also exposed to air], add them both, and you'd need to also count the area of the upper rim four times the thickness \times the length of one side of the box.
In your example of a solid cube with solid cylinder removed [say drilled out from top to bottom]: Then you'd need to sum the surface area of the 6 sides, prior to drilling, and after drilling depending on the radius $r$ of the open circles, on top and bottom, you'd need to subtract $2 \times \pi r^2$ and then you'd need to add $2\pi rx,$ [where $x$ is the height of one of the sides,
The difference between the two is how we define a box, and whether we are to take it ideal [no thickness of the walls/sides of the box] which we must do unless provided with the added dimension of thickness of the walls.
What is the surface area of a rectangular prism with an open top?
Surface Area of a rectangular prism = 2 [lh +wh + lw ] Square units.
What is the formula for the surface area of a box?
To find the surface area of a cuboid which has 6 rectangular faces, add the areas of all 6 faces. Or, you can label the length [l], width [w], and height [h] of the cuboid and use the formula: surface area [SA]=2lw+2lh+2hw.