Answer:
Dimensions of the can should be r = 7.15 cm and height = 14.33 cm.
Step-by-step explanation:
A cylindrical can can hold the amount of water = 2.3 liters
Since 1 liter = 1000 cm³
Therefore, 2.3 liters = 2.3×1000 = 2300 cm³
Volume of a cylinder 'V' = πr²h
2300 = πr²h
[tex]h=\frac{2300}{\pi r^{2} }[/tex]---------(1)
We have to calculate the dimensions of the can that will minimize the cost.
Total surface area A = 2πrh + 2πr²
To minimize the cost of the material used we will find the derivative of the total surface area and equate the derivative to zero.
A = [tex]2\pi r(\frac{2300}{\pi r^{2}})+2\pi r^{2}[/tex]
A = [tex]\frac{4600}{r}+2\pi r^{2}[/tex]
A' = [tex]-\frac{4600}{r^{2}}\frac{dh}{dr}+4\pi r[/tex]
A' = 0
[tex]-\frac{4600}{r^{2}}\frac{dh}{dr}+4\pi r[/tex] = 0
[tex]-\frac{4600}{r^{2}}+4\pi r=0[/tex]
[tex]4\pi r=\frac{4600}{r^{2} }[/tex]
[tex]r^{3}=\frac{4600}{4\times \pi }[/tex]
[tex]r=\sqrt[3]{\frac{4600}{4\times 3.14} }[/tex]
[tex]r=\sqrt[3]{366.242}[/tex]
r = 7.15 cm
From equation (1)
[tex]h=\frac{2300}{\pi r^{2}}[/tex]
= [tex]\frac{2300}{\pi \times (7.15)^{2} }[/tex]
= 14.33 cm
Therefore, can having the radius = 7.15 cm and h = 14.33 cm will have the minimum cost.
The dimensions of the can that will minimize the cost is radius of 7.15 cm and height of 14.3 cm
Volume = πr²h
2.3 liters = 2300 cm³, hence:
2300 cm³ = πr²h
h = (2300 / πr²)
Surface area (A) = 2πrh + 2πr²
A = 2πr(2300 / πr²) + 2πr²
A = 4600/r + 2πr²
The cost is minimum at A' = 0, hence:
A' = -4600/r² + 4πr
-4600/r² + 4πr = 0
4600/r² = 4πr
4πr³ = 4600
r = 7.15 cm
h = 2300 / (π * 7.15²) = 14.3 cm
The dimensions of the can that will minimize the cost is radius of 7.15 cm and height of 14.3 cm
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