Answer:
$ 81.77
Step-by-step explanation:
Let the volume of the rectangular storage cylinder be V = lwh where l = length, w = width and h = height of container.
Now, since the container is open at the top, its bottom area is lw and its side area is 2lh + 2wh = 2(l + w)h
So, the cost for the base material is 5lw and that for the side material is 3 × 2(lh + wh) = 6(lh + wh).
The total cost C is thus C = 5lw + Â 6(l + w)h
We know that the length is twice the width, so. l = 2w and the volume = 10 m³. So, V = lwh = (2w)wh = 2w²h. Thus h = V/2w² = 10/2w² = 5/w².
Substituting l and h into C, we have
C = 5lw + Â 6(l + w)h
C = 5(2w)w +  6((2w) + w)(5/w²)
C = 10w² +  6(3w)(5/w²)
C = 10w² +  90/w
Now C is a function of w the width
C(w) = 10w² +  90/w
We differentiate C(w) and equate it to zero to find the turning point of C(w)
dC(w)/dt = d{10w² +  90/w)/dt
dC(w)/dt = 20w - 90/w²
equating it to zero, we have
dC(w)/dt = 0
20w - 90/w² = 0
20w = 90/w²
w³ = 90/20
w³ = 4.5
w = ∛4.5
We differentiate dC(w)/dt again and insert w to determine if this is a minum or maximum point. So,
d²C(w)/dt² = d(20w - 90/w²)/dt
d²C(w)/dt² = 20 + 180/w³
substituting w³ = 4.5, we have
d²C(w)/dt² = 20 + 180/4.5
d²C(w)/dt² = 20 + 40 = 60
Since d²C(w)/dt² = 60 > 0
w = ∛4.5 is a minimum for C(w).
So, substituting w = ∛4.5 into C(w), we have
C(w) = 10w² +  90/w
C(w) = 10(∛4.5)² +  90/∛4.5
C(w) = 10(1.65)² +  90/1.65
C(w) = 27.225 + 54.545
C(w) = 81.77
So the cost for the least expensive of such container is $ 81.77
The cost of materials for the least expensive such material is $81.77
The surface area of an open-lid rectangular container is:
[tex]\mathbf{A = 2 \times Sides + Base}[/tex]
So, we have:
[tex]\mathbf{A = 2(lh + wh) + lw}[/tex]
The base costs $5 per square meter, and the sides cost $3 per square meter.
So, the cost function is:
[tex]\mathbf{C = 3 \times 2(lh + wh) + 5 \times lw}[/tex]
[tex]\mathbf{C = 6(lh + wh) + 5 lw}[/tex]
The length is twice the width.
This implies that:
[tex]\mathbf{l =2w}[/tex]
So, we have:
[tex]\mathbf{C = 6(2wh + wh) + 5 \times 2w^2}[/tex]
[tex]\mathbf{C = 6(3wh) + 10w^2}[/tex]
[tex]\mathbf{C = 18wh + 10w^2}[/tex]
The volume is calculated as:
[tex]\mathbf{V = lwh}[/tex]
Substitute 10 for V and 2w for l
[tex]\mathbf{2w^2h = 10}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{5}{w^2}}[/tex]
Substitute [tex]\mathbf{h = \frac{5}{w^2}}[/tex] in [tex]\mathbf{C = 18wh + 10w^2}[/tex]
[tex]\mathbf{C = 18w \times \frac{5}{w^2} + 10w^2}[/tex]
[tex]\mathbf{C = \frac{90}{w} + 10w^2}[/tex]
Differentiate
[tex]\mathbf{C' = -\frac{90}{w^2} + 20w}[/tex]
Set to 0
[tex]\mathbf{0 = -\frac{90}{w^2} + 20w}[/tex]
Rewrite as:
[tex]\mathbf{20w = \frac{90}{w^2} }[/tex]
Multiply both sides by w^2
[tex]\mathbf{20w^3 = 90 }[/tex]
Divide both sides by 20
[tex]\mathbf{w^3 = 4.5 }[/tex]
Take cube roots of both sides
[tex]\mathbf{w = 1.65 }[/tex]
Substitute 1.65 for w in [tex]\mathbf{C = \frac{90}{w} + 10w^2}[/tex]
[tex]\mathbf{C = \frac{90}{1.65} + 10 \times 1.65^2}[/tex]
[tex]\mathbf{C = 81.77}[/tex]
Hence, the cost of materials for the least expensive such material is $81.77
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