Show that the rectangle of maximum perimeter which can be inscri

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 Multiple Choice QuestionsShort 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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 Multiple Choice QuestionsLong Answer Type

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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.


Let O be the centre of circle of radius a. Let ABCD be the rectangle inscribed in the circle such that AB = x, AD = y
Now,     AB squared plus BC squared space equals space AC squared
therefore space space space space space space straight x squared plus straight y squared space equals space 4 straight a squared
therefore space space space space space space space space space space space space space space space straight y squared space equals space 4 straight a squared minus straight x squared
therefore space space space space space space space space space space space straight y space equals space square root of 4 straight a squared minus straight x squared end root space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis


Let P be the perimeter of rectangle
         therefore space space space straight P space equals 2 straight x plus 2 straight y       
         therefore space space straight P space equals 2 straight x plus 2 square root of 4 straight a squared minus straight x squared end root                                                  open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
          therefore space space space space dP over dx equals 2 plus 2 open parentheses fraction numerator negative 2 straight x over denominator 2 square root of 4 straight a squared minus straight x squared end root end fraction close parentheses
therefore space space space space space space dP over dx space equals space 2 minus fraction numerator 2 straight x over denominator square root of 4 straight a squared minus straight x squared end root end fraction
space space space space space space space space dP over dx space equals space 0 space space space space gives space us space space space 2 minus fraction numerator 2 straight x over denominator square root of 4 straight a squared minus straight x squared end root end fraction space equals 0
rightwards double arrow space space space straight x space equals space square root of 4 straight a squared minus straight x squared end root space space space space space space space rightwards double arrow space space space space straight x squared space equals space 4 straight a squared minus straight x squared space space space space rightwards double arrow space space space space 2 straight x squared space equals space 4 straight a squared space space space space space space rightwards double arrow space space space space straight x squared space equals space 2 straight a squared

space space space space space
rightwards double arrow space space space straight x space equals space straight a square root of 2 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 space straight x space cannot space be space negative close square brackets.
                  fraction numerator straight d squared straight P over denominator dx squared end fraction space space equals space 0 minus 2 space open square brackets fraction numerator square root of 4 straight a squared minus straight x squared end root. space 1 minus straight x. space open parentheses begin display style fraction numerator negative 2 straight x over denominator 2 square root of 4 straight a squared minus straight x squared end root end fraction end style close parentheses over denominator left parenthesis square root of 4 straight a squared minus straight x squared end root right parenthesis squared end fraction close square brackets

                 equals negative 2 open square brackets fraction numerator left parenthesis 4 straight a squared minus straight x squared right parenthesis space plus straight x squared over denominator left parenthesis 4 straight a squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space equals space minus fraction numerator 8 straight a squared over denominator left parenthesis 4 straight a squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction

When space straight x space equals space straight a square root of 2. space space fraction numerator straight d squared straight P over denominator dx squared end fraction space equals space minus fraction numerator 8 straight a squared over denominator left parenthesis 4 straight a squared minus 2 straight a squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction space equals space minus fraction numerator 8 straight a squared over denominator 2 square root of 2 straight a cubed end fraction space equals space minus fraction numerator 2 square root of 2 over denominator straight a end fraction less than 0
therefore space space space straight x space equals space straight y
therefore space space space straight P space is space maximum space when space rectangle space becomes space straight a space square.
Hence space the space result.
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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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 Multiple Choice QuestionsShort 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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 Multiple Choice QuestionsLong 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.

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 Multiple Choice QuestionsShort 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.
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