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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.

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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 $1 third straight h$

Let PR = h be the height of the right circular cone and PQ = y be the height of right circular cylinder of radius v.
In rt.

angle straight d space RQB comma space space RQ over QB space equals space cot space straight alpha

therefore space space space space space space space RQ over straight x space equals space cot space straight alpha comma space space space where space QB space equals space straight x space space space space space space rightwards double arrow space space space space space RQ space space equals space straight x space cot space straight alpha therefore space space space space space space space space space space straight y space equals space PR space minus space QR space equals space straight h minus straight x space cot space straight alpha 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 right circular cylinder.

therefore space space space space straight v space equals space πx squared left parenthesis straight h minus straight x space cot space straight alpha right parenthesis

open square brackets Volume space space equals space πr squared straight h close square brackets

dv over dx space equals space straight pi open square brackets straight x squared straight d over dx left parenthesis straight h minus straight x space cot space straight alpha right parenthesis space plus space left parenthesis straight h minus straight x space cot space straight alpha right parenthesis space straight d over dx left parenthesis straight x squared right parenthesis close square brackets

equals space straight pi open square brackets straight x squared left parenthesis negative cot space straight alpha right parenthesis space plus space left parenthesis straight h minus straight x space cot space straight alpha right parenthesis thin space left parenthesis 2 space straight x right parenthesis close square brackets space equals space πx open square brackets negative straight x space cot space straight alpha space plus space 2 space straight h space minus space 2 straight x space cot space straight alpha close square brackets

dv over dx space equals space πx left square bracket 2 straight h space minus space 3 space straight x space cot space straight alpha right square bracket

Now

dv over dx space equals space 0 space space space space gives space us space space space space straight pi space straight x left square bracket 2 straight h minus space 3 space straight x space cot space straight alpha right square bracket space equals 0

therefore space space space space 2 straight h minus 3 straight x space cot space straight alpha space equals space 0 space space as space space straight x space not equal to 0 space space space space space rightwards double arrow space space space space straight x space equals space fraction numerator 2 space straight h over denominator 3 space cot space straight alpha end fraction space equals space fraction numerator 2 straight h over denominator 3 end fraction space tan space space straight alpha space space space space space space space space fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi open square brackets straight x space left parenthesis negative 3 space cot space straight alpha right parenthesis space plus space 2 straight h space minus space 3 space straight x space cot space straight alpha close square brackets

space equals space straight pi left square bracket negative 3 straight x space cot space straight alpha space plus space 2 straight h space minus space 3 straight x space cot space straight alpha right square bracket space equals space straight pi left square bracket 2 straight h minus 6 straight x space cot space straight alpha right square bracket

When

straight x space equals space fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha comma space space then space

fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi space open square brackets 2 straight h minus 6. space fraction numerator 2 straight h over denominator 3 end fraction tan space straight alpha space cot space straight alpha close square brackets space equals space straight pi left square bracket 2 straight h minus 4 straight h right square bracket space equals space minus 2 πh less than 0

therefore space space space space space space straight v space is space maximum space when space straight x space equals space fraction numerator 2 straight h over denominator 3 end fraction space tan space space straight alpha

therefore space space space space space space straight v space equals space straight h minus fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha. space space cot space straight alpha

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

equals space straight h minus fraction numerator 2 straight h over denominator 3 end fraction space equals space straight h over 3.

Hence the result.

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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.

84 Views

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.

96 Views

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

79 Views