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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.

76 Views

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.

94 Views

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.

Let h be the height and r be the base radius of the inscribed cylinder in a sphere of radius R.
In rt. angled $increment OCA comma$
$OC squared plus CA squared space space equals OA squared therefore space space space space straight h squared over 4 plus straight r squared space equals space straight R squared therefore space space space space space straight r squared space equals space straight R squared minus straight h squared over 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Let V be the volume of the 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 minus straight h cubed right parenthesis space space space space space space space dV over dh space equals space straight pi over 4 left parenthesis 4 straight R squared minus 3 straight h squared right parenthesis and 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 straight h right parenthesis space equals space minus fraction numerator 3 πh over denominator 2 end fraction For space straight V space to space be space maxima space or space minima. space dV over dx space equals space 0 therefore 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 space rightwards double arrow space space space space space space space space space 4 straight R squared minus 3 straight h squared space equals space 0 space space space space space space space space rightwards double arrow space space space space space space straight h squared space equals space fraction numerator 4 straight R squared over denominator 3 end fraction therefore space space space straight h space equals space fraction numerator 2 straight R over denominator square root of 3 end fraction When space straight h space equals space fraction numerator 2 straight R over denominator square root of 3 end fraction 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 space 0 therefore space space space space space straight V 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 Max space volume space equals space straight pi over 4 open parentheses 4 straight R squared cross times fraction numerator 2 over denominator square root of 3 end fraction straight R. 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 cross times fraction numerator 8 straight R cubed over denominator 3 square root of 3 end fraction left parenthesis 3 minus 1 right parenthesis space equals fraction numerator 4 πR cubed over denominator 3 square root of 3 end fraction$

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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 $4 over 81 πh cubed$

121 Views

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 $4 over 27 πh cubed space tan squared straight alpha.$

86 Views

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 $5 square root of 3 space cm space is space 500 space straight pi space cm cubed.$