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Margaret [11]
3 years ago
5

A rectangular 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 b

ottom is $15 and for the top and the sides is $10, what is the minimum cost of the constructed box?

Mathematics
1 answer:
mixer [17]3 years ago
8 0

Answer:

  $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.

_____

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).

__

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. <em>Cost is minimized when the side length of the base is 8 ft</em>.

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a

The estimate is  - 0.0265\le  K \le  0.0465

b

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Step-by-step explanation:

The number of microchips broken in method A is  n_1 = 400

The number of faulty breaks of method A is  X_1 = 32

 The number of microchips broken in method B is  n_2  = 400

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  The proportion of the faulty breaks to the total breaks in method A is

       p_1 = \frac{32}{400}

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 The proportion of the faulty to the total breaks in method B is

      p_2 =  \frac{28}{400}

     p_2 =  0.07

For this estimation the standard error is  

      SE =  \sqrt{ \frac{p_1 (1 - p_1)}{n_1}  + \frac{p_2 (1- p_2 )}{n_2} } }

  substituting values

       SE =  \sqrt{ \frac{0.08 (1 - 0.08)}{400}  + \frac{0.07 (1- 0.07 )}{400} } }

      SE = 0.0186

The z-values of confidence coefficient of 0.95 from the z-table is  

       z_{0.95} =  1.96

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as

        K = [p_1 - p_2 ] \pm z_{0.95} * SE

substituting values

        K = [0.08 - 0.07 ] \pm 1.96 *0.0186

         K  =  - 0.0265 \ or  \ K  =  0.0465

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is  

      - 0.0265\le  K \le  0.0465

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