Show that the maximum volume of the cylinder which can be inscri

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Application of Derivatives

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.

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340.
Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius $5 square root of 3 space cm space is space 500 space straight pi space cm cubed.$

Let h be the height and R be the base radius of the inscribed cylinder.
r, radius of sphere =

5 square root of 3 space cm

In

increment OAC comma

OC squared plus CA squared space equals space OA squared space space space space space rightwards double arrow space space space space space space open parentheses straight h over 2 close parentheses squared plus space straight R squared space equals space left parenthesis 5 square root of 3 right parenthesis squared therefore space space space space straight R squared space equals space 75 space minus space 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... left parenthesis 1 right parenthesis

Let V be volume of the cylinder.

therefore space space space space space space space straight V space equals space πR squared straight h space equals space straight pi space open parentheses 75 minus straight h squared over 4 close parentheses straight h space equals space straight pi over 4 left parenthesis 300 minus straight h squared right parenthesis space straight h space equals space straight pi over 4 left parenthesis 300 minus straight h cubed right parenthesis space space space space space space dV over dh space equals space straight pi over 4 left parenthesis 300 minus 3 straight h squared right parenthesis. space space space space space space space dV over dh space equals space 0 space space space space space space space space space space rightwards double arrow space space space space space straight pi over 4 left parenthesis 300 minus 3 straight h squared right parenthesis space equals space 0 space space space space space space rightwards double arrow space space 300 minus 3 straight h squared space equals space 0 space space rightwards double arrow space space space space space space space straight h squared space equals space 100 space space space space space space space space space space space space rightwards double arrow space space space space space straight h space equals space 10 space space space space space space space space space space 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 negative 6 space straight h right parenthesis space equals negative fraction numerator 3 πh over denominator 2 end fraction

When space space space space space space space straight h space equals space 10 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 cross times 10 space equals space minus 15 space straight pi space less than 0 therefore space space space space straight V space is space maximum space when space straight h space equals space 10 therefore space space space space space greatest space volume space equals space straight pi over 4 left parenthesis 300 minus 100 right parenthesis thin space left parenthesis 10 right parenthesis space equals space straight pi over 4 cross times 200 space cross times space 10 space equals space 500 space straight pi space cm cubed.

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