Answer:
base side = 20cm
height = 40 cm
minimum cost = $7200
Step-by-step explanation:
Let x (cm) be the base side and h (cm) be the height of the box. Since the volume of the box must be 16000 cubic cm, we have:
[tex]x^2h = 16000[/tex]
[tex]h = 16000/x^2[/tex]
The areas of the top and bottom is
[tex]A_B = 2x^2[/tex]
And so their cost is:
[tex]C_B = A_T*3 = 6x^2[/tex]
The areas of the sides is
[tex]A_S = 4xh[/tex]
And so their cost is:
[tex]C_S = A_S*1.5 = 6xh[/tex]
We can substitute [tex]h = 16000/x^2[/tex] into the equation above to get
[tex]C_S = 96000/x[/tex]
So the total cost of the bases and the sides:
[tex]C = C_S + C_B = 6x^2 + 96000/x[/tex]
This is the function we would want to minimize. We can start by taking the 1st derivative, and set it to 0
[tex]C^{'} = 12x - 96000/x^2 = 0[/tex]
[tex]12x = 96000/x^2[/tex]
[tex]x^3 = 96000/12 = 8000[/tex]
[tex]x = 20cm[/tex]
So the height would be
[tex]h = 16000/x^2 = 16000/20^2 = 40 cm[/tex]
So the minimum total cost would be
[tex]C = 6x^2 + 96000/x = 6*20^2 + 96000/20 = 7200[/tex]
The dimensions that minimize the cost of the box are length and width of 20 cm, and a height of 40 cm, and the minimum cost is $7200
The surface area of an closed box is:
[tex]\mathbf{A = 2lw + 2hw + 2lh}[/tex]
The base and top costs $3, and the sides cost $1.5.
So, the cost function is:
[tex]\mathbf{C = 3 \times 2lw + 1.5 \times (2hw + 2lh)}[/tex]
[tex]\mathbf{C = 6lw + 3hw + 3lh}[/tex]
The box is square based.
So: l = w
This gives
[tex]\mathbf{C = 6l^2 + 3lh + 3lh}[/tex]
[tex]\mathbf{C = 6l^2 + 6lh}[/tex]
The volume is calculated as:
[tex]\mathbf{V =l^2h}[/tex]
So, we have:
[tex]\mathbf{l^2h = 16000}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{16000}{l^2}}[/tex]
Substitute [tex]\mathbf{h = \frac{16000}{l^2}}[/tex] in [tex]\mathbf{C = 6l^2 + 6lh}[/tex]
[tex]\mathbf{C = 6l^2 + 6l \times \frac{16000}{l^2}}[/tex]
[tex]\mathbf{C = 6l^2 + \frac{96000}{l}}[/tex]
Differentiate
[tex]\mathbf{C' = 12l - \frac{96000}{l^2}}[/tex]
Set to 0
[tex]\mathbf{12l - \frac{96000}{l^2} = 0}[/tex]
Rewrite as:
[tex]\mathbf{\frac{96000}{l^2} = 12l}[/tex]
Multiply both sides by l^2
[tex]\mathbf{12l^3 = 96000}[/tex]
Divide both sides by 12
[tex]\mathbf{l^3 = 8000}[/tex]
Take cube roots
[tex]\mathbf{l = 20}[/tex]
Recall that:
[tex]\mathbf{h = \frac{16000}{l^2}}[/tex]
So, we have:
[tex]\mathbf{h = \frac{16000}{20^2}}[/tex]
[tex]\mathbf{h = 40}[/tex]
Recall that:
[tex]\mathbf{C = 6l^2 + \frac{96000}{l}}[/tex]
So, we have:
[tex]\mathbf{C = 6 \times 20^2 + \frac{96000}{20}}[/tex]
[tex]\mathbf{C = 7200}[/tex]
Hence, the dimensions that minimize the cost of the box are length and width of 20 cm, and a height of 40 cm, and the minimum cost is $7200
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