Given Volume = 4000 cm^3 = (x^2)
So y = 4000/(x^2)
Substitute the y into the area function.
Area = (x^2) + 4x [ 4000 / (x^2) ]
Area = (x^2) + 16 000/x
I use differentiation method to find the minimum dimension of the box.
Let y = (x^2) + 16 000/x
dy/dx = 2x + (- 16 000 x^ -2)
dy/dx = 2x - 16 000/(x^2)
Differentiate dy/dx again, d^2y/dx^2 = 2 + 32 000 x^ -3
So Second differentiation = 2 + 32 000/(x^3)
When dy/dx = 0, 2x - 16 000/(x^2) = 0
2x^3 -16000 = 0
2X^3 = 16000
x^3 = 8000
x = 20
When x = 20, second differentiation = 2 + 32 000/(2000^3)
second differentiation = 6 (positive means this is minimum)
Sometimes we get 2 answers for x, that is why we use second differentiation to determine which x is maximum or minimum.
Therefore, when x = 20, y = 4000/20^2
y = 10
So length/width = 20 cm, height = 10 cm.