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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 ?

92 Views

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

82 Views

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 $4 over 81 πh cubed$

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

angle straight d space increment space RQB comma space space space RQ over QB space equals space cot space 30 degree

therefore space space space space space space RQ over straight x space equals space cot space 30 degree comma space space where space QB space equals space straight x rightwards double arrow space space space space space space space RQ space space equals straight x space cot space 30 degree therefore space space space space space space space space RQ space equals space square root of 3 space straight x therefore space space space space space space space space space space space space space space straight y space equals space PR space minus space QR space space equals space straight h minus square root of 3 space straight x 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 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 square root of 3 space straight x right parenthesis

open square brackets because space space volume space space equals space πr squared straight h close square brackets

dv over dx space equals straight pi open square brackets straight x squared left parenthesis negative square root of 3 right parenthesis space plus space left parenthesis straight h minus square root of 3 straight x right parenthesis space 2 straight x close square brackets space equals space straight pi space open square brackets negative square root of 3 straight x squared plus 2 hx minus 2 square root of 3 straight x squared close square brackets space space space space space space space space space space equals space straight pi open square brackets 2 hx minus 3 square root of 3 space straight x squared close square brackets space space equals space straight pi space straight x left parenthesis 2 straight h minus 3 square root of 3 straight x right parenthesis Now space dv over dx space equals 0 space space gives space us space space space πx left parenthesis 2 straight h minus 3 square root of 3 space straight x right parenthesis space equals space 0 therefore space space space space space 2 straight h minus 3 square root of 3 space straight x space equals space 0 space space space space space as space space straight x not equal to 0 space space space space space space rightwards double arrow space space space straight x space equals space fraction numerator 2 straight h over denominator 3 square root of 3 end fraction fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi left parenthesis 2 straight h minus 6 square root of 3 space straight x right parenthesis

When space straight x space equals space fraction numerator 2 straight h over denominator square root of 3 end fraction. space fraction numerator straight d squared straight v over denominator dx squared end fraction space equals straight pi space open parentheses 2 straight h minus 6 square root of 3. space fraction numerator 2 straight h over denominator 3 square root of 3 end fraction close parentheses space equals space straight pi space left parenthesis 2 straight h minus 4 straight h right parenthesis space space space space space space space space space space space space space space space space space space equals negative 2 πh space less than space 0 therefore 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 square root of 3 end fraction Max. space value space of space straight v space equals space straight pi space open parentheses fraction numerator 2 straight h over denominator 3 square root of 3 end fraction close parentheses squared space space space open parentheses straight h minus square root of 3 cross times fraction numerator 2 over denominator 3 square root of 3 end fraction close parentheses space equals space straight pi space cross times space fraction numerator 4 straight h squared over denominator 27 end fraction cross times straight h over 3 equals space space fraction numerator 4 πh cubed over denominator 81 end fraction

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