An open tank with a square base and vertical sides is to be cons

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 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.

101 Views

 Multiple Choice QuestionsLong 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.

90 Views

323.

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

599 Views

 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?

83 Views

Advertisement
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.

86 Views

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.

82 Views

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.

88 Views

 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.

362 Views

Advertisement
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

 Multiple Choice QuestionsShort Answer Type

Advertisement

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.


Let x be sides of the square base and h be the depth of the given tank of volume V.
therefore space space space space straight V space equals space straight x squared straight h space space space space space space space rightwards double arrow space space space space straight h space space equals space space straight V over straight x squared                                                    ...(1)
Surface area of tanki = left parenthesis straight x squared plus 4 xh right parenthesis space sq. space units
Let Rs. p be the cost per square unit of material and C be the total cost.
therefore space space space straight C space equals space left parenthesis straight x squared plus 4 xh right parenthesis straight p space equals space open parentheses straight x squared plus fraction numerator 4 straight V over denominator straight x end fraction close parentheses space straight p. space where space straight p space is space constant. space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
space space space space space dC over dx equals straight p open parentheses 2 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction close parentheses
Now comma space dC over dx equals 0 space space gives space us
space space

          straight p open parentheses 2 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction close parentheses space equals 0 space space space space space space or space space space space 2 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction space equals 0
therefore space space space straight x cubed minus 2 straight V space equals space 0 space space space space space space space rightwards double arrow space space space straight x cubed space equals space 2 thin space straight V                                                ...(2)
space fraction numerator straight d squared straight C over denominator dx squared end fraction space equals space straight p open parentheses 2 plus fraction numerator 8 space straight V over denominator straight x cubed end fraction close parentheses space space equals straight p space open parentheses 2 plus 8 over straight x cubed. space straight x cubed over 2 close parentheses space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 2 right parenthesis close square brackets
space space space space space space space space space space space equals space straight p left parenthesis 2 plus 4 right parenthesis space equals space 6 straight p greater than 0
therefore space space space space space space straight C space is space minimum.
Now space space space straight h space equals space straight V over straight x squared space equals space fraction numerator begin display style straight x cubed over 2 end style over denominator straight x squared end fraction space equals space straight x over 2
therefore space space space depth space of space tank space equals space half space of space its space width.
923 Views

Advertisement
Advertisement