a box with a square base and open top must have a volume of 4000cm^3. What dimensions of the box would minimize the amount of material used?

Question

I get that you need to find the derivative of something to get the minimum value, but how do you get that original equation? and then what do you do after getting the derivative of it?

Answer 1

V = volume = 4000 = s^2(h) where s= side of the square bottom and h= height of the box

h = 4000/s^2

A = area of the bottom plus 4 sides' areas = s^2 + (4)(sh) = s^2 + 4(4000/s^2)(s) = s^2 + 16000/s

A' = 2s - 16000/s^2 = 0

divide by 2

s - 8000/s^2 = 0

multiply by s^2

s^3 -8000=0

s^3 = 8000

s= the cube root of 8,000 = (8000)^(1/3)

s = 20 cm = side of the square bottom

20x20x20 = 8000

h=4000/20^2 = 4000/400 = 10 cm high

dimensions that minimize amount of material are: 10 cm by 20 cm by 20 cm where 10 cm is the height of the box and 20 cm is the length of a side of the square bottom

the original equation you need is the sum of the area of the bottom plus the areas of each vertical side, but make it a function of just one variable, either the side or the height. A(s) = s^2 + 4hs, replace h with its equivalent in terms of s. h=4000/s^2. That gives A(s) = s^2 +4(4000/s^2)s = s^2 +16000/s

take the derivative of A(s) and set it equal to zero. A'(s) = 0, then solve for s