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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.

Let x, x, y be the length, width and height of the cuboid such that base is a square of side x.
$therefore space space space straight V space equals space straight x squared straight y$ ...(1)
where V is volume of cuboid.
Let S be surface area of cuboid.
$therefore space space space straight S space equals space straight x squared plus straight x squared plus 4 xy space space space open square brackets because space space straight S space equals area space of space base space plus space area space of space top space plus area space of space four space walls close square brackets space space space space space space space space space space space$
$space equals space 2 straight x squared plus 4 straight x. space straight V over straight x squared$ $open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets$
$therefore space space space space straight S space equals space 2 straight x squared plus fraction numerator 4 straight V over denominator straight x end fraction space space space space rightwards double arrow space space space dS over dx space equals 4 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction$
$dS over dx space equals 0 space space space space rightwards double arrow space space 4 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction space equals space 0 space space space space space rightwards double arrow space space space straight x cubed space equals space straight V space space space space rightwards double arrow space space straight x space equals space straight V to the power of 1 third end exponent fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space 4 plus fraction numerator 8 straight V over denominator straight x cubed end fraction$

When $straight x space equals straight V to the power of 1 third end exponent. space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space 4 plus fraction numerator 8 straight V over denominator straight V end fraction space equals space 4 plus 8 space equals space 12 space greater than 0$
$therefore$ S is minimum when $straight x equals straight V to the power of 1 third end exponent$
When $straight x space equals space straight V to the power of 1 third end exponent comma space space space space space straight y space equals space straight V over straight x squared space equals space straight V over straight V to the power of begin display style 2 over 3 end style end exponent space equals space straight V to the power of 1 third end exponent space equals space straight x$
$therefore space space space space space straight S space is space minimum space when space length comma space width space and space height space are space straight x space each therefore space space space space straight S space is space minimum space when space cuboid space is space straight a space cube. space$

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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?

83 Views

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.

257 Views

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