A rectangular box is to have a square base and a volume of 63 ft3. If the material for the base costs $0.22/ft2, the material for the sides costs $0.09/ft2, and the material for the top costs $0.20/ft2, determine the dimensions (in ft) of the box that can be constructed at minimum cost.

9514 1404 393

   3 ft square base, 7 feet high

Step-by-step explanation:

For minimum cost, each pair of opposite sides costs the same as any other pair of opposite sides.

If s is the edge length of the base, then the areas of the top and bottom are s² and their total cost is ...

  (s²)(0.22 +0.20) = 0.42s² . . . . cost of top and bottom

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For a volume of 63 ft³, the height of the side of the box is ...

  63/s² = h

and the area of one side of the box is ...

  hs = 63/s

A pair of opposite sides will have a cost of ...

  (63/s)(0.09 +0.09) = 11.34/s . . . . cost of a pair of opposite sides

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We want these two costs to be the same, so we have ...

  0.42s² = 11.34/s

  s³ = 11.34/0.42 = 27

  s = ∛27 = 3 . . . . feet

  h = 63/s² = 7 . . . . feet

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The box dimensions are 3 feet square by 7 feet tall.

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Additional comment

Each pair of opposite sides will cost $3.78, so the total cost of the box is $11.34.

For the non-believers, we have shown a graph of the total cost of the box as a function of the base dimension (s). It has a minimum at $11.34 when s=3. You can find that by differentiating the cost function and and finding the zero(s) of that (or by reading the values from the graph).

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

the correct answer is $7.38

step-by-step explanation: