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Application of Derivatives

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Long Answer Type

331. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m^3. If building of tank costs Rs. 70 per sq. metres for the base and Rs. 45 per square metre for sides. What is the cost of least expensive tank ?

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Short Answer Type

332.
Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.

Let r be the radius of base of circular cylinder and h be its height. Let V be the volume and S be the total surface area.

therefore space space 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 rightwards double arrow space space space straight h space equals straight V over πr squared

...(1)
Also,

straight S space equals space 2 πrh plus 2 πr squared space equals space 2 πr. space straight V over πr squared plus 2 πr squared

open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

therefore space space straight S space equals fraction numerator 2 space straight V over denominator straight r end fraction plus 2 πr squared therefore space space space dS over dr space equals negative fraction numerator 2 straight V over denominator straight r squared end fraction plus 4 πr Now space dS over dr equals 0 space space space rightwards double arrow space space space space fraction numerator negative 2 straight V over denominator straight r squared end fraction plus 4 space straight pi space straight r space equals space 0 space space space space space space space space rightwards double arrow space space space space 2 space straight V space equals space 4 πr cubed

rightwards double arrow space space space 2 πr squared straight h space equals space 4 πr cubed

open square brackets because space space space straight V space equals space πr squared straight h close square brackets

rightwards double arrow space space space straight r space equals space straight h over 2

Now,

fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space fraction numerator 4 straight V over denominator straight r squared end fraction plus 4 straight pi

When space straight r space equals straight h over 2 comma space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space fraction numerator 4 straight V over denominator straight h cubed end fraction cross times 8 plus 4 straight pi space equals space fraction numerator 32 space straight V over denominator straight h cubed end fraction plus 4 straight pi space greater than 0 therefore space space space straight S space is space minimum space when space straight r space equals space straight h over 2 space straight i. straight e. comma space straight h space equals space 2 straight r space straight i. straight e. comma space height space equals space diameter.

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Long Answer Type

333.

Prove that the radius of the right circular cylinder of greatest curved surface which can be inscribed in a given cone is half of that of the cone.

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Short Answer Type

334. Show that the height of the right circular cylinder of maximum volume that can be inscribed in a given circular cone of height h is

1 third straight h

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Long Answer Type

335.

Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is a sphere of radius R is $fraction numerator 2 straight R over denominator square root of 3 end fraction.$ Also, find the maximum volume.

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Short Answer Type

336.

Find the height of a right circular cylinder of maximum volume, which can be inscribed in a sphere of radius 9 cm.

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337.

Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle 30° is $4 over 81 πh cubed$

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Long Answer Type

338.

Show that height of the cylinder of greatest volume which can be inscribed in a right ciruclar cone of height h and semi vertical angle $straight alpha$ is one-third that of the cone and the greatest volume of cylinder is $4 over 27 πh cubed space tan squared straight alpha.$

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Short Answer Type

339. Find the volume of the largest cylinder that can be inscribed in a sphere of radius r.

75 Views

340.

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius $5 square root of 3 space cm space is space 500 space straight pi space cm cubed.$