Show that the height of the cylinder, open at the top, of given

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 Multiple Choice QuestionsShort Answer Type

321.

The combined resistance R of two resistors R1 and R2 (R1 , R2 > 0) is given by 1 over straight R space equals space 1 over straight R subscript 1 plus 1 over straight R subscript 2.
If R1 + R2 = C (a constant), show that the maximum resistance R is obtained by choosing R1 = R2.

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 Multiple Choice QuestionsLong Answer Type

322.

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius is a square of side straight a square root of 2.

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323.

Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

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 Multiple Choice QuestionsShort Answer Type

324.

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area?
Or
Of all the closed cylindrical tin cans (right circular) which enclose a given volume of 100 cubic cm., which has the minimum surface area?

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325.

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

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326.

Show that a cylinder of given volume open at the top has minimum total surface area provided its height is equal to the radius of its base.

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327.

Show that the height of the cylinder, open at the top, of given surface area and greatest volume is equal to the radius of its base.


Let r be the radius of base of circular cylinder and h be its height. Let V be the volume and S be total surface area.
therefore space space straight S space equals space πr squared plus 2 πrh space rightwards double arrow space space space straight h space equals space fraction numerator straight S minus πr squared over denominator 2 πr end fraction                        ...(1)
       straight V equals space πr squared straight h space equals space πr squared space open parentheses fraction numerator straight S minus πr squared over denominator 2 πr end fraction close parentheses                                  open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
therefore space space space straight V space equals space 1 half left parenthesis Sr minus πr cubed right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space dV over dr equals space 1 half left parenthesis straight S minus 3 πr squared right parenthesis
space space space space space dV over dr space equals 0 space space space space space rightwards double arrow space space space space 1 half left parenthesis straight S space minus space 3 πr squared right parenthesis space equals space 0 space space space space rightwards double arrow space space space straight r space equals space square root of fraction numerator straight S over denominator 3 straight pi end fraction end root
space space space space space fraction numerator straight d squared straight V over denominator dr squared end fraction space equals space 1 half left parenthesis 0 minus 6 πr right parenthesis space space equals space minus 3 πr
For space space space space straight r space equals space square root of fraction numerator straight S over denominator 3 straight pi end fraction end root comma space fraction numerator straight d squared straight v over denominator dr squared end fraction space equals space minus 3 straight pi square root of fraction numerator straight S over denominator 3 straight pi end fraction end root space equals space minus square root of 3 πS end root less than 0
therefore space space space space space straight V space is space greatest space when space straight r space equals space square root of fraction numerator straight S over denominator 3 straight pi end fraction end root     
From (1),  straight h equals fraction numerator straight S minus straight pi. space begin display style fraction numerator straight S over denominator 3 straight pi end fraction end style over denominator 2 straight pi. space square root of begin display style fraction numerator straight S over denominator 3 straight pi end fraction end style end root end fraction space equals space fraction numerator begin display style fraction numerator 2 straight S over denominator 3 end fraction end style over denominator 2 square root of begin display style πS over 3 end style end root end fraction space equals fraction numerator 2 straight S over denominator 3 end fraction cross times fraction numerator square root of 3 over denominator square root of πS end fraction space equals space square root of fraction numerator straight S over denominator 3 straight pi end fraction end root
therefore height = radius of base         


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 Multiple Choice QuestionsLong Answer Type

328.

A rectangle is inscribed in a semi-circle of radius r with one of its sides on the diameter of the semi-circle. Find the dimensions of the rectangle so that its area is maximum Find also this area.

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329.

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

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 Multiple Choice QuestionsShort Answer Type

330. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water, show that the cost of the material will be least when the depth of the tank is half of its width.
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