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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.

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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.

Let x be the radius of the base of the cone of height PR = h. Let r be the radius of cylinder of height PQ = y.
Now

increment RQA space space and space space increment RPB

are similar

therefore space space space QA over PB space equals RQ over RP rightwards double arrow space space space space straight r over straight x space equals space fraction numerator straight h minus straight y over denominator straight h end fraction rightwards double arrow space space space space space straight r space equals space fraction numerator left parenthesis straight h minus straight y right parenthesis space straight x over denominator straight h end fraction

Let S be the curved surface of cylinder

therefore space space straight S space equals space 2 πry space equals 2 straight pi fraction numerator left parenthesis straight h minus straight y right parenthesis space straight x over denominator straight h end fraction straight y space equals fraction numerator 2 πx over denominator straight h end fraction left parenthesis hy minus straight y squared right parenthesis therefore space space dS over dv space equals space fraction numerator 2 πx over denominator straight h end fraction left parenthesis straight h minus 2 straight y right parenthesis

For S to be maximum or minimum.

dS over dy space equals space 0

therefore space space space space space space fraction numerator 2 πx over denominator straight h end fraction left parenthesis straight h minus 2 straight y right parenthesis space equals space 0 space space space space space rightwards double arrow space space space space space straight h minus 2 straight y space equals space 0 space space space space space space space rightwards double arrow space space space straight h space equals space 2 straight y Also comma space space fraction numerator straight d squared straight S over denominator dy squared end fraction space equals fraction numerator 2 πx over denominator straight h end fraction left parenthesis negative 2 right parenthesis space equals space fraction numerator negative 4 πx over denominator straight h end fraction When space straight h space equals space 2 straight y comma space space space space space fraction numerator straight d squared straight s over denominator dy squared end fraction space equals negative fraction numerator 4 πx over denominator straight h end fraction less than 0 therefore space space space space space straight S space is space maximum space when space straight h space equals space 2 straight y Now comma space space space space straight r space equals space fraction numerator straight x over denominator 2 straight y end fraction left parenthesis 2 straight y minus straight y right parenthesis space equals space straight x over 2

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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.

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340.

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