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Assignment work

A cylinder with an open top has a capacity of 2m³ and is made from sheet steel. Neglecting any overlaps at the joints, find the dimensions of the cylinder so that the amount of sheet steel used is minimum

can anyone help its for my HNC assignment on differentation

Reply 1

Ed5

If it has a capacity of 2m³ then

\pi r^2h = 2 \Rightarrow h = \frac{2}{\pi r^2}

To minimise the amount of material, you need to minimise the surface area used. As it's open top, that means the surface area will be the curved edge and the circular bottom:

A = \pi r^2 + 2\pi r h

Substitution from the first formula gives

A =\pi r^2 + 2\pi r \cdot \frac{2}{\pi r^2} = \pi r^2 + \frac{4}{r}

To find the minimum or maximum of an equation, you set its differential equal to zero and then verify whether it is a min or max by finding the second differential (>0 is min, <0 is max). Hence:

\frac{dA}{dr} = 2\pi r - 4r^{-2} = 0 \rightarrow 2\pi r^3 - 4 = 0 \rightarrow r^3 = \frac{2}{\pi} \rightarrow r \approx 0.860m

\frac{d^2A}{dr^2} = 2\pi + 8r^{-3} \approx 11.4

- this is above zero which confirms it's the minimum!

Using

h = \frac{2}{\pi r^2}

h \approx 0.860

(edited 7 years ago)

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