Respuesta :
Answer:
Length = 2 ft
Width = 2 ft
Height = 5 ft
Step-by-step explanation:
Let the square base of the box has one side = x ft
Therefore, area of the base = x² ft²
Cost of the material to prepare the base = $0.35 per square feet
Cost to prepare the base = $0.35x²
Let the height of the box = y ft
Then the volume of the box = x²y ft³ = 20
[tex]y=\frac{20}{x^{3} }[/tex] -----(1)
Cost of the material for the sides = $0.10 per square feet
Area of the sides = 4xy
Cost to prepare the sides of the box = $0.10 × 4xy
= $0.40xy
Cost of the material to prepare the top = $0.15 per square feet
Cost to prepare the top = $0.15x²
Total cost of the box = 0.35x² + 0.40xy + 0.15x²
From equation (1),
Total cost [tex]C=0.35x^{2}+(0.40x)\times \frac{20}{x^{2} }+0.15x^{2}[/tex]
[tex]C=0.35x^{2}+\frac{8}{x}+0.15x^{2}[/tex]
[tex]C=0.5x^{2}+\frac{8}{x}[/tex]
Now we take the derivative of C with respect to x and equate it to zero,
[tex]C'=0.5(2x)-\frac{8}{x^{2}}[/tex] = 0
[tex]x-\frac{8}{x^{2}}=0[/tex]
[tex]x=\frac{8}{x^{2} }[/tex]
[tex]x^{3}=8[/tex]
x = 2 ft.
From equation (1),
[tex](2)^{2}y=20[/tex]
4y = 20
y = 5 ft
Therefore, Length and width of the box should be 2 ft and height of the box should be 5 ft for the minimum cost to construct the rectangular box.
The volume of a box is the amount of space in it.
The dimensions that minimize cost are:
[tex]\mathbf{Length = 2ft}[/tex]
[tex]\mathbf{Width = 2ft}[/tex]
[tex]\mathbf{Height= 5ft}[/tex]
The volume of the box is:
[tex]\mathbf{V = 20}[/tex]
Assume the dimension of the base is x, and the height is h.
The volume of the box will be:
[tex]\mathbf{V = x^2h}[/tex]
So, we have:
[tex]\mathbf{x^2h = 20}[/tex]
The area of the sides is:
[tex]\mathbf{A_1 = 4xh}[/tex]
The cost of the side material is 0.10.
So, the cost is:
[tex]\mathbf{C_1 = 0.10 \times 4xh}[/tex]
[tex]\mathbf{C_1 = 0.4xh}[/tex]
The area of the base is:
[tex]\mathbf{A_2 = x^2}[/tex]
The cost of the base material is 0.35.
So, the cost is:
[tex]\mathbf{C_2 = 0.35x^2}[/tex]
The area of the top is:
[tex]\mathbf{A_3 = x^2}[/tex]
The cost of the top material is 0.15.
So, the cost is:
[tex]\mathbf{C_3 = 0.15x^2}[/tex]
So, the total cost is:
[tex]\mathbf{C = 0.4xh + 0.35x^2 + 0.15x^2}[/tex]
[tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]
Recall that: [tex]\mathbf{x^2h = 20}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{20}{x^2}}[/tex]
Substitute [tex]\mathbf{h = \frac{20}{x^2}}[/tex] in [tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]
[tex]\mathbf{C = 0.4x \times \frac{20}{x^2} + 0.5x^2}[/tex]
[tex]\mathbf{C = \frac{8}{x} + 0.5x^2}[/tex]
Differentiate
[tex]\mathbf{C' = -\frac{8}{x^2} + x}[/tex]
Set to 0
[tex]\mathbf{-\frac{8}{x^2} + x = 0}[/tex]
Rewrite as:
[tex]\mathbf{x = \frac{8}{x^2}}[/tex]
Multiply both sides by x^2
[tex]\mathbf{x^3 = 8}[/tex]
Take cube roots of both sides
[tex]\mathbf{x = 2}[/tex]
Recall that:
[tex]\mathbf{h = \frac{20}{x^2}}[/tex]
So, we have:
[tex]\mathbf{h = \frac{20}{2^2}}[/tex]
[tex]\mathbf{h = 5}[/tex]
Hence, the dimensions that minimize cost are:
[tex]\mathbf{Length = 2ft}[/tex]
[tex]\mathbf{Width = 2ft}[/tex]
[tex]\mathbf{Height= 5ft}[/tex]
Read more about volumes at:
https://brainly.com/question/13101997
