It is desired to construct a cylindrical vessel of capacity 500 cubic metres open at the top.What should be the dimensions of the vessel so that the material need is minimum, given that the thickness of the material used is 2 cm.

Question

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Answer 1

Let r and h be the internal radius and height of the cylindrical vessel and let V denote the volume of the material.
Now, $V = π r^{2} h$
$\Rightarrow 500 = π r^{2} h$
$\Rightarrow h = \frac{5 0 0}{π r ^{2}}$ ...(1)

Now, $S = (2 π r h + π r^{2}) \frac{2}{1 0 0}$ (As $\frac{2}{1 0 0}$ is thickness)
$= (\frac{1 0 0 0}{r} + π r^{2}) . \frac{1}{5 0}$
For maximum or minimum ,
$\frac{d s}{d r} = (- \frac{1 0 0 0}{r ^{2}} + 2 π r) \frac{1}{5 0} = 0$
$\Rightarrow π r^{3} = 500$
$∴ π r^{3} = 500 = V = π r^{2} h$
or $r = (\frac{5 0 0}{π})^{1 / 3} = h$

Clearly $\frac{d ^{2} S}{d r ^{2}} = (\frac{4 V}{r ^{3}} + 2 π) . \frac{1}{5 0} = + i v e$
$∴$ For minimum volume, $r = (\frac{5 0 0}{π})^{1 / 3} = h$

Answer 2

Let r and h be the internal radius and height of the cylindrical vessel and let V denote the volume of the material.

Now,

V = π r^{2} h

\Rightarrow 500 = π r^{2} h

\Rightarrow h = \frac{5 0 0}{π r ^{2}}

...(1)

Now,

S = (2 π r h + π r^{2}) \frac{2}{1 0 0}

(As

\frac{2}{1 0 0}

is thickness)

= (\frac{1 0 0 0}{r} + π r^{2}) . \frac{1}{5 0}

For maximum or minimum ,

\frac{d s}{d r} = (- \frac{1 0 0 0}{r ^{2}} + 2 π r) \frac{1}{5 0} = 0

\Rightarrow π r^{3} = 500

∴ π r^{3} = 500 = V = π r^{2} h

or

r = (\frac{5 0 0}{π})^{1 / 3} = h

Clearly

\frac{d ^{2} S}{d r ^{2}} = (\frac{4 V}{r ^{3}} + 2 π) . \frac{1}{5 0} = + i v e

∴

For minimum volume,

r = (\frac{5 0 0}{π})^{1 / 3} = h