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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.

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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.

Let h be the height and R be the base radius of the inscribed cylinder.
In

increment OCA comma

OC squared plus CA squared space equals space OA squared space space space space rightwards double arrow space space space space open parentheses straight h over 2 close parentheses squared plus straight R squared space equals space straight r squared therefore space space space space space straight R squared space equals space straight r squared minus straight h squared over 4 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 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 space space space space space space space... left parenthesis 1 right parenthesis

Let V be volume of cylinder

therefore space space space straight V space equals space πR squared straight h space equals space straight pi open parentheses straight r squared minus straight h squared over 4 close parentheses straight h space equals space straight pi over 4 left parenthesis 4 straight r squared straight h minus straight h cubed right parenthesis space space space space space space space space dV over dh space equals straight pi over 4 left parenthesis 4 straight r squared minus 3 straight h squared right parenthesis space space space space space space space space dV over dh space equals space 0 space space space space rightwards double arrow space space space space space straight pi over 4 left parenthesis 4 straight r squared minus 3 straight h squared right parenthesis space equals space 0 space space space space space rightwards double arrow space space space 4 straight r squared space minus 3 space straight h squared space equals space 0 space space space space space space rightwards double arrow space space space space space space straight h squared space equals space 4 over 3 straight r squared space space space space space space space space space space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space straight pi over 4 left parenthesis 0 minus 6 straight h right parenthesis space equals space minus fraction numerator 3 πh over denominator 2 end fraction When space straight h space equals space fraction numerator 2 over denominator square root of 3 end fraction straight r comma space space space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space minus fraction numerator 3 straight pi over denominator 2 end fraction. space fraction numerator 2 straight r over denominator square root of 3 end fraction space equals space minus square root of 3 space straight pi space straight r space less than 0

therefore space space space space space space space straight V space space is space maximum space when space straight h space equals space fraction numerator 2 over denominator square root of 3 end fraction straight r therefore space space space space greatest space volume space space equals space straight pi over 4 open parentheses 4 straight r squared. space space fraction numerator 2 over denominator square root of 3 end fraction straight r space minus space fraction numerator 8 over denominator 3 square root of 3 end fraction straight r cubed close parentheses space space space space equals space straight pi over 4 open parentheses fraction numerator 8 over denominator square root of 3 end fraction straight r cubed space minus space fraction numerator 8 over denominator 3 square root of 3 end fraction straight r cubed close parentheses space equals space straight pi over 4 open parentheses fraction numerator 24 straight r cubed minus 8 straight r cubed over denominator 3 square root of 3 end fraction close parentheses equals straight pi over 4 open parentheses fraction numerator 16 straight r cubed over denominator 3 square root of 3 end fraction close parentheses space equals space fraction numerator 4 πr cubed over denominator 3 square root of 3 end fraction

75 Views

340.

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

79 Views