For a given current surface area of right circular cone when the

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 Multiple Choice QuestionsLong Answer Type

341. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 over 27 of the volume of the sphere.
Or
Find the volume of the largest cone that can be inscribed in a sphere of radius r.

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342.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan to the power of negative 1 end exponent square root of 2.

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343.

For a given current surface area of right circular cone when the volume is maximum. Prove that the semi-vertical angle is straight theta where sin space straight theta space equals space fraction numerator 1 over denominator square root of 3 end fraction.


Let x be radius, h be the vertical height and y be the slant height of cone.
∴    y2 = h2+ x2 ...(1)
Let V be volume of cone.


therefore space space space space space straight V space equals space 1 third πx squared straight h space equals space 1 third πx squared square root of straight y squared minus straight x squared end root space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
therefore space space space space space straight V squared space equals space 1 over 9 straight pi squared straight x to the power of 4 left parenthesis straight y squared minus straight x squared right parenthesis
therefore space space space space space space fraction numerator 9 straight V squared over denominator straight pi squared end fraction space equals space straight x to the power of 4 left parenthesis straight y squared minus straight x squared right parenthesis 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 space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Now total surface area of cone = constant
therefore space space space space πx squared plus πxy space equals constant
rightwards double arrow space space space space straight x squared plus xy space equals space constant space space rightwards double arrow space space space 2 straight x plus straight x dy over dx plus straight y. space 1 space equals space 0
rightwards double arrow space space space space space dy over dx space equals space minus open parentheses fraction numerator 2 straight x plus straight y over denominator straight x end fraction close parentheses 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 space space space space space space space space space space space... left parenthesis 3 right parenthesis
Now, V will be maximum when straight V squared space space or space space fraction numerator 9 straight V squared over denominator straight pi squared end fraction is maximum.
Let space fraction numerator 9 straight V squared over denominator straight pi squared end fraction equals straight Z
therefore space space space space space straight z space equals space straight x to the power of 4 left parenthesis straight y squared minus straight x squared right parenthesis 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 space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 2 right parenthesis close square brackets space space space
space space space space space space space space dz over dx space equals space 4 straight x cubed space. space left parenthesis straight y squared minus straight x squared right parenthesis space plus space straight x to the power of 4 open parentheses 2 straight y dy over dx minus 2 straight x close parentheses
space space space space space space space space space space space space space equals space 4 straight x cubed left parenthesis straight y squared minus straight x squared right parenthesis plus straight x to the power of 4 open square brackets 2 straight y open parentheses negative fraction numerator 2 straight x plus straight y over denominator straight x end fraction close parentheses minus 2 straight x close square brackets 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 open square brackets because space of space left parenthesis 3 right parenthesis close square brackets
            equals 4 straight x cubed left parenthesis straight y squared minus straight x squared right parenthesis space minus space 2 straight x cubed straight y left parenthesis 2 straight x plus straight y right parenthesis space minus 2 straight x to the power of 5 space equals space 4 straight x cubed straight y squared minus 4 straight x to the power of 5 minus 4 straight x to the power of 4 straight y minus 2 straight x cubed straight y squared space minus 2 straight x to the power of 5
equals space 2 straight x cubed straight y squared minus 6 straight x to the power of 5 minus 4 straight x to the power of 4 straight y

   Now comma space dz over dx space equals space 0 space space gives space us space space space 2 straight x cubed straight y cubed space minus space 6 straight x to the power of 5 space minus space 4 straight x to the power of 4 straight y space equals space 0
space space space space therefore space space space space straight x cubed left parenthesis straight y squared minus 3 straight x squared minus 2 xy right parenthesis space equals space 0 space space space rightwards double arrow space space space space straight x cubed left parenthesis straight y plus straight x right parenthesis thin space left parenthesis straight y minus 3 straight x right parenthesis space equals space 0
space space space space therefore space space space space straight x space equals space 0 comma space space minus straight y comma space space space straight y over 3
  Now space straight x space equals space 0 comma space minus straight y space are space not space possible comma space space space therefore space space space straight y space equals space 3 straight x space space
When space straight y space is space slightly space less than space 3 straight x comma space space dz over dx space equals plus ve
When space straight y space is space slightly greater than 3 straight x comma space space dz over dx space equals space minus ve
therefore space space space space space space space straight z space or space space straight V space is space maximum space for space straight y space equals space 3 straight x.
Let space space straight theta space be space the space semi minus vertical space angle
therefore space space space space space space sin space straight theta space equals space straight x over straight y space equals space fraction numerator straight x over denominator 3 straight x end fraction space equals space 1 third
space space space

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 Multiple Choice QuestionsShort Answer Type

344.

Prove that a conical tent of given capacity will require the least amount of convas when the height is square root of 2 times the radius of the base.

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345.

Show that the right circular cone of least curved surface and given volume has an altitude equal to an altitude equal to square root of 2 times the radius of the base.

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 Multiple Choice QuestionsLong Answer Type

346.

Find the altitude of a right circular cone of maximum curved surface which can be inscribed in a sphere of radius r.

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347.

Find the equation of the line through the point (3, 4) which cuts from the first quadrant a triangle of minimum area.

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348.

Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.

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349. Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.
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350.

Prove that the perimeter of a right-angled triangle of given hypotenuse equal to 5 cm is maximum when the triangle is isosceles.

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