You are constructing a cardboard box from a piece of cardboard with the dimensions 2 m by 4 m. You then cut equal-size squares from each corner so you may fold the edges. What are the dimensions (in m) of the box with the largest volume

Dimensions:

l = 3,15 m

w=1,15 m

x = 0,42 m (height)

V (max) = 1,52 m³

Step-by-step explanation:

The cardboard is L = 4 m and W = 2 m

Let call x the length of the square to cut in each corner, then, volume of open box is:

For the side L is L - 2*x l = 4 - 2*x

For the side W is W - 2*x w = 2 - 2*x

The height is x

Volume of the open box, as function of x is:

V (x) = (4 - 2x) * (2 - 2x) * x ⇒ V (x) = (8 - 8x - 4x + 4x²) * x

V (x) = (8 - 12x + 4x²) * x V (x) = 8x - 12x² + 4x³

V (x) = 8x - 12x² + 4x³

Taking derivatives on both sides of the equation

V' (x) = 8 - 24x + 12x²

V' (x) = 0 8 - 24x + 12x² = 0 reordering 12x² - 24x + 8 = 0

or 3x² - 6x + 2 = 0

A second degree equation. Solving for x

x₁,₂ = (6 ± √36 - 24) / 6

x₁,₂ = (6 ± 3.46) / 6

x₁ = 6 + 3,46 / 6 x₁ = 1.58 we dismiss such solution because 1,58 * 2 = 3,15 and is bigger than 2 one of the side of the cardboard

x₂ = (6 - 3,46) / 6

x₂ = 0,42 m

Dimensions of the open box

l = 4 - 2*x l = 4 - 0,85 l = 3,15 m

w = 2 - 2*x w = 2 - 0,85 w = 1,15 m

x = 0,42 m

V (max) = 3,15*1,15*0,42

V (max) = 1,52 m³