Answer:
The dimension that minimizes the container is width = 1; base = 2 and height = 5
The minimum cost is $36
Explanation:
Let the width be x
So:
[tex]Width = x[/tex]
[tex]Base = 2 * Width[/tex]
[tex]Base = 2 * x[/tex]
[tex]Base = 2x[/tex]
[tex]Height = h[/tex]
Volume of the box is:
[tex]Volume = 10m^3[/tex] ---- Given
Volume is calculated as:
[tex]Volume = Base * Width * Height[/tex]
[tex]Volume = 2x * x * h[/tex]
[tex]Volume = 2x^2h[/tex]
Substitute 10 for Volume
[tex]10 = 2x^2h[/tex]
Make h the subject
[tex]h = \frac{10}{2x^2}[/tex]
[tex]h = \frac{5}{x^2}[/tex]
Next, we calculate area of the sides.
[tex]Area = Base + Sides[/tex]
Because it has an open top, the area is:
[tex]Base\ Area = Base * Width[/tex]
[tex]Sides\ Area = 2[(Width * Height) + (Base * Height)][/tex]
[tex]Base\ Area = 2x * x[/tex]
[tex]Base\ Area = 2x^2[/tex]
[tex]Sides\ Area = 2[(Width * Height) + (Base * Height)][/tex]
[tex]Side\ Area = 2[(x * h) + (2x * h)][/tex]
[tex]Side\ Area = 2[(xh) + (2xh)][/tex]
[tex]Side\ Area = 2[3xh][/tex]
[tex]Side\ Area = 6xh[/tex]
The base area costs $6 per m²
So, the cost of 2x² would be:
[tex]Cost = 6 * 2x^2[/tex]
[tex]Cost = 12x^2[/tex]
The side cost area costs $0.8 per m²
So, 6xh would cost
[tex]Cost = 0.8 * 6xh[/tex]
[tex]Cost = 4.8xh[/tex]
Total Cost (C) is:
[tex]C = 12x^2 + 4.8xh[/tex]
Recall that [tex]h = \frac{5}{x^2}[/tex]
So:
[tex]C = 12x^2 + 4.8x *\frac{5}{x^2}[/tex]
[tex]C = 12x^2 + 4.8 *\frac{5}{x}[/tex]
[tex]C = 12x^2 + \frac{4.8 *5}{x}[/tex]
[tex]C = 12x^2 + \frac{24}{x}[/tex]
Take derivative of C
[tex]C^{-1} = 24x - \frac{24}{x^2}[/tex]
Take LCM
[tex]C^{-1} = \frac{24x^3 - 24}{x^2}[/tex]
Equate [tex]C^{-1}[/tex] to 0
[tex]0 = \frac{24x^3 - 24}{x^2}[/tex]
Cross multiply
[tex]24x^3 - 24 = 0 * x^2[/tex]
[tex]24x^3 - 24 = 0[/tex]
Add 24 to both sides
[tex]24x^3 = 24[/tex]
Divide through by 24
[tex]x^3 = 1[/tex]
Take cube roots of both sides
[tex]x = 1[/tex]
Recall that
[tex]Width = x[/tex]
[tex]Base = 2x[/tex]
[tex]Height = h[/tex] and [tex]h = \frac{5}{x^2}[/tex]
Solve for these dimensions:
[tex]Width = 1[/tex]
[tex]Base = 2 * 1[/tex]
[tex]Base = 2[/tex]
[tex]h = \frac{5}{1^2}[/tex]
[tex]h = \frac{5}{1}[/tex]
[tex]h = 5[/tex]
i.e.
[tex]Height = 5[/tex]
Hence, the dimension that minimizes the container is width = 1; base = 2 and height = 5
Recall that
[tex]C = 12x^2 + \frac{24}{x}[/tex]
Substitute 1 for x
[tex]C = 12(1^2) + \frac{24}{1}[/tex]
[tex]C = 12 + 24[/tex]
[tex]C = 36[/tex]
The minimum cost is $36
