# In a shipping container only boxes with height 1 m can be kept. A person has 8 m^2 of cardboard, what is the volume of the largest box that he can make that can be kept in the container.

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Only boxes with a height of 1 m can be placed in the shipping container. The person has 8 m^2 of cardboard to make a box. The height of the box is 1 m. If the length of the box is L and the width is W, W*L*2 + 2*W*1 + 2*L*1 = 8.

The volume of the box is V = W*L*1

W*L*2 + 2*W*1 + 2*L*1 = 8

=> W*L*2 + 2*W = 8 - 2*L

=> W(2L + 2) = (8 - 2*L)

=> W = `(8 - 2*L)/(2L + 2)`

=> W = `(4 - L)/(L + 1)`

The volume of the box is V = `((4 - L)/(L + 1))*L*1`

To maximize V, solve `(dV)/(dL) = 0`

=> `((4 - L)/(L+1)) + L*(-1*(L+1) - (4 - L))/(L+1)^2 = 0`

=> `(4 - L)(L + 1) + L*(-1*(L+1) - (4 - L)) = 0`

=> `4L - L^2 + 4 - L - L^2 - L - 4L + L^2 = 0`

=> `4 - 2L - L^2 = 0`

=> L = `sqrt 5 - 1`

W = `sqrt 5 - 1`

The width of the box is `sqrt 5 - 1`

The maximum volume of the box is `(sqrt 5 -1)^2*1 = 6-2*sqrt5`

**The maximum volume of the box is `6 - 2*sqrt 5` m^3**