Show that the semi-vertical angle of the cone of the maximum vol

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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.


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


therefore space space space space straight V space equals space 1 third πx squared straight h space equals space 1 third straight pi left parenthesis straight y squared minus straight h squared right parenthesis space straight h space space space space space space space space space space space space space space left square bracket because space space of space left parenthesis 1 right parenthesis right square bracket
therefore space space space space space space straight V space equals space 1 third straight pi left parenthesis straight y squared straight h minus straight h cubed right parenthesis
space space space space space space dV over dh space equals straight pi over 3 left parenthesis straight y squared minus 3 straight h squared right parenthesis
For V to be maximum or minimum,  dV over dh space equals space 0
therefore space space space straight pi over 3 left parenthesis straight y squared minus 3 straight h squared right parenthesis space equals space 0 space space space space space space space rightwards double arrow space space space space space straight y squared minus 3 straight h squared space equals space 0 space space space space space rightwards double arrow space space space space straight h space equals space fraction numerator straight y over denominator square root of 3 end fraction
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 3 left parenthesis 0 minus 6 straight h right parenthesis space equals space minus 2 space straight pi space straight h
At space space space straight h equals fraction numerator straight y over denominator square root of 3 end fraction space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space minus 2 straight pi fraction numerator straight y over denominator square root of 3 end fraction less than 0
therefore space space space space space space space straight V space is space max. space when space straight h space equals space fraction numerator straight y over denominator square root of 3 end fraction space straight i. straight e. comma space space space space when space straight h squared space equals space straight y squared over 3
straight i. straight e. comma space when space 3 straight h squared space equals space straight y squared space space space space space space space space straight i. straight e. comma space space space when space 3 straight h squared space equals space straight x squared plus straight h squared 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 1 right parenthesis close square brackets
straight i. straight e. comma space when space 2 straight h squared space equals space straight x squared space space space space space space space space straight i. straight e. space space space space when space square root of 2 space straight h space equals space straight x
straight i. straight e. comma space when space straight x over straight h equals space square root of 2 space space space space space space space straight i. straight e. space space space when space tanθ space equals square root of 2 space space space space straight i. straight e. comma space when space tanθ space equals space 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.

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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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