Of all the closed cylindrical cans (right circular), of a 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

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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?


Let V be the volume of cylinder whose radius is r; height is h and surface area is S.
therefore space space space space space space straight V space equals space πr squared straight h comma space space space space or space space space space 100 space equals space πr squared straight h
therefore space space space space space space straight h equals space 100 over πr squared                                                                      ...(1)
Now, straight S space equals space 2 πr squared space plus space 2 πrh space equals space 2 πr squared plus 2 πr. space 100 over 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 equals space 2 πr squared plus 200 over straight r
space space space space space space space dS over dr space equals space 4 πr minus 200 over straight r squared
Now space space dS over dr space equals 0 space space give space us
space space space space space space 4 πr minus 200 over straight r squared space equals space 0 comma space space space space space rightwards double arrow space space 4 πr cubed minus 200 space equals space 0 space space space space space rightwards double arrow space space straight r cubed space equals space 50 over straight pi space space space rightwards double arrow space space space space straight r space equals space open parentheses 50 over straight pi close parentheses to the power of 1 third end exponent
space space space space space space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space 4 straight pi space plus 400 over straight r cubed
space space space space space space space space space

space
When space space straight r space equals space open parentheses 50 over straight pi close parentheses to the power of 1 third end exponent comma space space space space space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space 4 straight pi plus 400 over 50 cross times straight pi space equals space 12 straight pi greater than 0
therefore space space space space straight S space is space minimum space when space straight r space equals space open parentheses 50 over straight pi close parentheses to the power of 1 third end exponent 
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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.

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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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