Abox of volume 36 m3 with square bottom and no top is constructed out of two different materials. the cost of the bottom is $40/m2 and the cost of the sides is $30/m2 . find the dimensions of the box that minimize total cost. (let s denote the length of the side of the square bottom of the box and h denote the height of the box.)

Step-by-step explanation:

The volume of the box is given by the formula ...

  V = Bh

where B is the area of the bottom, and h is the height. We are given values for V and B, so we can write ...

  36 = s²·h

  h = 36/s²

Then the cost of materials for the box is ...

  bottom cost = 40s²

  one side cost = 30sh = 30s(36/s²) = 1080/s

  4-side cost = 4 × one side cost = 4×1080/s = 4320/s

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  Total cost = bottom cost + 4-side cost

  c = 40s² +4320/s

This will be at a minimum when its derivative with respect to s is zero:

  dc/ds = 0 = 80s -4320/s²

  54 = s³

  s = 3∛2 ≈ 3.77976 . . . . meters

  h = 36/s² = 36s/s³ = (2/3)s ≈ 2.51984 . . . . meters

The dimensions of the box that minimize total cost are ...

  s = 3.78 m

  h = 2.52 m

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Please note that these dimensions make the cost of the box bottom be exactly the same as the cost of a pair of opposite sides. This is true of the optimal cost for any such problem. If the box has a top, then the total cost of top+bottom is the same as the total cost of any pair of opposite sides.

Since the box bottom is $40/m² and a pair of opposite sides total $60/m², we know the area of the side will be 40/60 times the area of the bottom. That is, the box will be 2/3 of a cube, and the bottom dimension will be ...

  ∛(3/2·36) = ∛54 . . . . meters

as we found above.

yes

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step-by-step explanation: