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
