JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Application of Derivatives

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A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of

V mm^{3}

, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container.

If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of

\frac{V}{250 π}

Answer: 4

Solution

Let

r

be the inner radius and

h

be the inner height. The inner volume is

V = π r^{2} h

, so

h = \frac{V}{π r ^{2}}

.

The outer radius is

r + 2

. The volume of material is:

V_{1} = π (r + 2)^{2} \cdot 2 + π [(r + 2)^{2} - r^{2}] h = 2 π (r + 2)^{2} + π h (4 r + 4)

= 2 π (r + 2)^{2} + 4 π (r + 1) h = 2 π [(r + 2)^{2} + \frac{2 ( r + 1 ) V}{π r ^{2}}]

For minimum, differentiate with respect to

r

\frac{d V _{1}}{d r} = 2 π [2 (r + 2) + \frac{2 V}{π} (- \frac{1}{r ^{2}} - \frac{2}{r ^{3}})] = 0

r = 10

2 (12) + \frac{2 V}{π} (- \frac{1}{100} - \frac{2}{1000}) = 0

24 - \frac{2 V}{π} \cdot \frac{12}{1000} = 0

V = \frac{24 \times 1000 π}{24} = 1000 π

Therefore,

\frac{V}{250 π} = \frac{1000 π}{250 π} = 4

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer

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JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Application of Derivatives

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