Show that a cylinder of given volume open at the top has minimum

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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.


Let r be the radius of base of circular cylinder and h be its height. Let V be volume and S be the total surface area.
therefore space space space straight V space equals space πr squared straight h space space space space space space space space space space space space space space space rightwards double arrow space space space space straight h space equals space straight V over πr squared                                          ...(1)
Also,    straight S space equals 2 πrh plus πr squared                                 open square brackets because space cylinder space is space open space at space top close square brackets
                equals 2 πr. space straight V over πr squared plus πr squared                                                             open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
therefore space space space space straight S space space equals fraction numerator 2 straight V over denominator straight r end fraction plus πr squared space space space space space space space space space space space space space rightwards double arrow space space space dS over dr space equals negative fraction numerator 2 straight V over denominator straight r squared end fraction plus 2 πr
Now,  dS over dr space equals 0 space space space space space rightwards double arrow space space space minus fraction numerator 2 straight V over denominator straight r squared end fraction plus 2 πr space equals space 0 space space space space rightwards double arrow space space space space 2 straight V space equals space 2 πr cubed
rightwards double arrow space space space space 2 πr squared straight h space equals space 2 πr cubed space space space space space space space space space space space space space space rightwards double arrow space space space straight r space equals space straight h space space space space space space space space space space space space space space open square brackets because space space straight V space equals space πr squared straight h close square brackets
Now comma space space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals fraction numerator 4 straight V over denominator straight h cubed end fraction plus 2 straight pi
When space straight r space equals space straight h comma space space space space space space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals fraction numerator 4 straight V over denominator straight h cubed end fraction plus 2 straight pi greater than 0
therefore space space space space space space straight S space is space minimum space when space straight h space equals space straight r space straight i. straight e. space height space space equals space radius space of space 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.

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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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