Astorage box with a square base must have a volume of 80 cubic centimeters. the top and bottom cost $0.20 per square centimeter and the sides cost $0.10 per square centimeter. find the dimensions that will minimize cost.

The dimensions that will minimize cost are 3.42 cm and 6.84 cm

Step-by-step explanation:

* Lets explain how to solve this problem

- We have a storage box with a square base

- The volume of the box is 80 cm³

* From the information above we can find relation between the two

 dimensions of the box

∵ The base of the box is a square with side length L cm

∵ The height of the box is H cm

∵ The volume of the box = area of its base × its height

- The base is a square and area the square = L² cm²

∴ The volume of the box = L² × H

∵ The volume of the box = 80 cm³

∴ L² × H = 80

- Lets find H in terms of L by divide both sides by L²

∴ H = 80/L² ⇒ (1)

- The cost of the top and bottom is $0.20 per cm²

- We can find the cost of top and bottom by multiplying the area of

  them by the cost per cm²

∵ The top and the bottom are squares with side length L cm

∴ The area of them = 2 × L² = 2L² cm²

∵ The cost per cm² is $0.20

∴ The cost of top and bottom = 2L² × 0.20 = 0.40L² ⇒ (2)

- Now we can find the cost of the lateral area (area of the 4 side faces)

 by multiplying the area of them by the cost per cm²

∵ The lateral area = the perimeter of its base × its height

∵ The base is a square with side length L cm

∴ The perimeter of the base = 4 × L = 4L cm

∵ The height of the box is H cm

∴ The lateral area = 4L × H

- Now lets replace H by L using equation (1)

∴ The lateral area = 4L × 80/L²

- To simplify it : 4 × 80 = 320 and L/L² = 1/L

∴ The lateral area = 320/L cm²

∵ The cost of the sides is $0.10 per cm²

∴ The cost of the lateral area = 320/L × 0.10 = 32/L ⇒ (3)

- Now lets find the total cost of the box by adding (2) and (3)

∴ The total cost (C) = 0.40L² + 32/L

* For the minimize cost we will differentiate the equation of the

  cost C with respect to the dimension L (dC/dL) and equate it

  by 0 to find the value of L which makes the cost minimum

- In differentiation we multiply the coefficient of L by its power and

 subtract 1 from the power

∵ C = 0.40L² + 32/L

- Lets change 32/L to 32L^(-1) ⇒ (we change the sign of the power by

 reciprocal it)

∴ C = 0.40L² + 32L^(-1)

-  Lets differentiate

∴ dC/dL = (0.40 × 2)L^(2 - 1) + (32 × -1)L^(-1 - 1)

∴ dC/dL = 0.80L - 32L^(-2)

- For the minimum cost put dC/dL = 0

∴ 0.80L - 32L^(-2) = 0 ⇒ add 32L^(-1) to both sides

∴ 0.80L = 32L^(-2)

- Change 32L^(-2) to 32/L² (we change the sign of the power by

 reciprocal it)

∴ 0.80L = 32/L² ⇒ use cross multiplication to solve it

∴ L³ = 32/0.80 = 40 ⇒ take ∛ for both sides

∴ L = ∛40 = 3.41995 ≅ 3.42 cm ⇒ to the nearest 2 decimal place

- Substitute this value of L in equation (1) to find H

∵ H = 80/L²

∴ H = 80/(∛40)² = 6.8399 ≅ 6.84 cm ⇒ to the nearest 2 decimal place

* The dimensions that will minimize cost are 3.42 cm and 6.84 cm

square pyramid

step-by-step explanation:

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d

step-by-step explanation: