Let the length of the base be 2x and the breadth of the base be x (since the length is twice the breadth).
The height of the box can be denoted by h.
The volume of the box is given as 288 cm³, so we have:
Volume of box = Length x Breadth x Height
288 = 2x * x * h
288 = 2x²h
h = 144/x²
To find the minimum amount of material used, we need to minimize the surface area of the box. The surface area of the box is given by:
Surface area = 2lh + 2bw + lb
Substituting the value of h, we get:
Surface area = 2(2x)(h) + 2(x)(2x) + (2x)(h)
Surface area = 4x(144/x²) + 4x² + 2x(144/x²)
Surface area = 576/x + 4x² + 288/x
To minimize the surface area, we need to differentiate it with respect to x, and set the derivative equal to zero:
d(Surface area)/dx = -576/x² + 8x + 288/x² = 0
Multiplying both sides by x², we get:
-576 + 8x³ + 288 = 0
Simplifying further, we get:
x³ = 36
Taking the cube root of both sides, we get:
x = 3∛(36)
x = 3∛(2^2 * 3^2)
x = 6∛2
Therefore, the breadth of the base is 6∛2 cm, and the length of the base is 2 times the breadth, which is 12∛2 cm.
Substituting the value of x in the equation for h, we get:
h = 144/(6∛2)²
h = 8∛2
Therefore, the dimensions of the box that uses the minimum amount of material are:
- Length: 12∛2 cm
- Breadth: 6∛2 cm
- Height: 8∛2 cm.