Box, w/ square base + open top w/ volume 32,000cm^3. Dimensions of box that min. amt. of material used.?

Question

Would love an elaborate solution. Optimization problem, college level calculus course. Thank you.

Answer 1

Call base length b and height h. Volume = hb^2
The expression to be minimized is
2b^2 + 4bh = Area.
We want to work with only one variable. Pick h
2 x 32000/h + 4 x h (32000/h)^1/2 . Simplify"
64000 (h^-1) + 320 h^(1/2) sqrt(5). The mult by 80 is from the simplificaion of the radical.
We want to take d/dh of this and minimize the derivitive to solve for h.

-64000 (h^-2) + 320 (1/2 ) sqrt(5/ h) = 0

Then 64000 h^(-2) = 160 sqrt (5/h)
400 = sqrt(5) h^3/2
so h^3/2 = 180 appx. and h = 28 cm appx.
The base area is 32000/28= 1200 appx. So b= about 35 cm.
NOTE: The "real answer" is the cube root of the volume without all this song and dance. I think my approximations (no calculator) account for the difference.

Answer 2

To solve this min/max problem, we first require the formula for the volume of this box. Normally, the volume of a box is equal to length times width times height, but in this case, length = width. That makes the formula V = (x^2)h But V = 32, so 32 = (x^2)h We also require the surface area of this box. Since there is no top, the surface area should be A = (area of the bottom) + (area of the 4 walls) A = x^2 + 4xh But, from the first equation, we can solve for h. 32 = (x^2)h implies that h = 32/(x^2). Plug this value of h into our area equation, and we get A = x^2 + 4x[32/(x^2)] Simplifying, we get A = x^2 + 4(32/x) A = x^2 + 128/x Making into a single fraction, A = (x^3 + 128)/x Now that we have one variable, we will now declare this as our area function, A(x). A(x) = (x^3 + 128)/x To minimize the area, take the derivative A'(x) and then make it 0. Using the quotient rule, A'(x) = [ (3x^2)(x) - (x^3 + 128)(1) ] / (x^2) A'(x) = [ 3x^3 - x^3 - 128 ] / (x^2) A'(x) = [ 2x^3 - 128 ] / (x^2) A'(x) = 2(x^3 - 64) / (x^2) Make A'(x) = 0, 0 = 2(x^3 - 64) / (x^2) We can neglect the bottom, because x^2 is not equal to 0 (x must represent a dimension). 0 = 2(x^3 - 64) 0 = x^3 - 64 64 = x^3 Therefore, x = 4 The minimum area occurs at x = 4. We can derive h, since h = 32/(x^2) h = 32/(4^2) h = 32/16 h = 2 The dimensions of the box that can be bult with minimum amount of materials would be 4" x 4" x 2" As a side note, the minimum surface area would be A(4), or A(4) = (4^3 + 128)/4 A(4) = (64 + 128)/4 A(4) = 192/4 A(4) = 48 square inches