let the sides of the base be x cm each
let the height be y
V= x^2 y
4000/x^2 = y
surface area = x^2 + 4xy
= x^2 + 4x(4000/x^2
= x^2 + 16000/x
d(surface area)/dx = 2x - 16000/x^2 = 0 for a min surface area
2x = 16000/x^2
x^3 = 16000
x = appr 25.2 cm
y = 4000/25.2^2 = appr 6.3 cm
A box with a square base and open top must have a volume of 4,000 cm3. Find the dimensions of the box that minimize the amount of material used.
2 answers
V= x^2 y
4000/x^2 = y
surface area = x^2 + 4xy
= x^2 + 4x(4000/x^2
= x^2 + 16000/x
d(surface area)/dx = 2x - 16000/x^2 = 0 for a min surface area
2x = 16000/x^2
x^3 = 8000
x = 20cm
y = 4000/400 = 10cm
4000/x^2 = y
surface area = x^2 + 4xy
= x^2 + 4x(4000/x^2
= x^2 + 16000/x
d(surface area)/dx = 2x - 16000/x^2 = 0 for a min surface area
2x = 16000/x^2
x^3 = 8000
x = 20cm
y = 4000/400 = 10cm