Arectangular box with a square base and a cover is to be built to contain 640 cubic feet. if the cost per square foot for the bottom is $15 and for the top and the sides is $10, what is the minimum cost of the constructed box?

  $4800

Step-by-step explanation:

The cost is minimized when any pair of opposite faces of the box costs the same as any other pair. Since the top+bottom total $25 per square foot and the sides (left+right or front+back) total $20 per square foot, the area of a side must be 25/20 = 5/4 times the area of the top or bottom.

If we let x represent the length of one side of the square bottom, then 5/4x will be the height of the box. Its volume will be ...

  (5/4x)(x²) = 640 ft³

  x³ = (4/5)(640 ft³) = 512 ft³ . . . . . . multiply by 4/5

  x = 8 ft . . . . . . . . . . . . . . . . . . . . . . . take the cube root

The base area is then (8 ft)² = 64 ft², and the cost of the top+bottom is ...

  64 ft² × $25/ft² = $1600

The three pairs of opposite sides will cost 3×$1600 = $4800.

The minimum cost of the constructed box is $4800.

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You can also write an equation for the total cost in terms of x, the side length of the base. That cost function will be ...

  c(x) = 15x² + 10x² + 10·4x(640/x²)

This simplifies to ...

  c(x) = 25x² +25600/x

A graph shows the minimum is c(8) = 4800 (as above).

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Algebraically, you can set the derivative to zero to find the value of x.

  dc/dx = 0 = 50x -25600/x²

  512 = x³ . . . . multiply by x²/50 and add 512

  8 = x . . . . . . . take the square root. Same solution as above. Cost is minimized when the side length of the base is 8 ft.