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Application of Derivatives

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Short Answer Type

321.

The combined resistance R of two resistors R₁ and R₂ (R₁ , R₂ > 0) is given by $1 over straight R space equals space 1 over straight R subscript 1 plus 1 over straight R subscript 2$ .
If R₁ + R₂ = C (a constant), show that the maximum resistance R is obtained by choosing R₁ = R₂.

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Long Answer Type

322.

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius is a square of side $straight a square root of 2.$

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323.

Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

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Short Answer Type

324.

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area?
Or
Of all the closed cylindrical tin cans (right circular) which enclose a given volume of 100 cubic cm., which has the minimum surface area?

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325.

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

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326.

Show that a cylinder of given volume open at the top has minimum total surface area provided its height is equal to the radius of its base.

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327.

Show that the height of the cylinder, open at the top, of given surface area and greatest volume is equal to the radius of its base.

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Long Answer Type

328.

A rectangle is inscribed in a semi-circle of radius r with one of its sides on the diameter of the semi-circle. Find the dimensions of the rectangle so that its area is maximum Find also this area.

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329.
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

Let R be radius and l be the slanted height of cone.
Let OD = x, OA = OB = OC = 12, where 12 is a radius of sphere.
$In space rt. space space angle straight d space increment ODC comma space space space space straight R space equals space square root of 144 minus straight x squared end root In space rt space angle straight d space increment ADC comma space space space space space space space space l space equals space square root of straight R squared plus left parenthesis straight x plus 12 right parenthesis squared end root space equals space square root of 144 minus straight x squared plus straight x squared plus 144 plus 24 straight x end root space space space space space space space space space equals square root of 288 plus 24 straight x end root space equals space square root of 24 space equals space square root of straight x plus 12 end root$


Let S be curved surface area of cone.
$therefore space space straight S space equals space πR l space equals space straight pi square root of 144 minus straight x squared end root space square root of 24 space square root of straight x plus 12 end root space equals space straight pi square root of 24 space left parenthesis straight x plus 12 right parenthesis thin space square root of 12 minus straight x end root space$
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 $equals space straight pi square root of 24 open square brackets fraction numerator 12 minus 3 straight x over denominator 2 square root of 12 minus straight x end root end fraction close square brackets$
For S to be maximum or minimum, $dS over dx equals 0$
$therefore space space space space straight pi square root of 24 space open square brackets fraction numerator 12 minus 3 straight x over denominator 2 square root of 12 minus straight x end root end fraction close square brackets space equals 0 space space space space rightwards double arrow space space space straight x equals space 4 When space straight x less than 4 space left parenthesis slightly right parenthesis comma space space dS over dx space equals space plus ve$
$and space when space straight x thin space greater than 4 space left parenthesis slightly right parenthesis comma space dS over dx space equals space minus ve$
$therefore space space at space straight x space equals space 4 comma space space space dS over dx space changes space from space plus ve space to space minus ve therefore space space straight S space is space maximum space at space straight x space equals space 4 therefore space space space space altitude space AD space equals space 4 plus 12 space equals space 16.$

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Short Answer Type

330. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water, show that the cost of the material will be least when the depth of the tank is half of its width.

923 Views