Show that among rectangles of given area, the square has the lea

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 Multiple Choice QuestionsShort Answer Type

281.

Divide  a number 15 into two parts such that the square of one multiplied with the cube of the other is a maximum.

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282.

Divide 15 into two parts such that the sum of squares is minimum.

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 Multiple Choice QuestionsLong Answer Type

283.

How should we choose two numbers, each greater than or equal to -2 whose  sum is 1 half so that the sum of the square of the first and cube of the second is the minimum?

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284.

The space s described in time t by a particle moving in a straight line is given by s = r5 – 40r + 30r2 + 80t – 250. Find the minimum value of its acceleration.

 

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 Multiple Choice QuestionsShort Answer Type

285.

Two sides of a triangle are given. Find the angle between them such that the area shall be maximum.

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286.

Show that of all the rectangles with a given perimeter, the square has the largest area.

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287.

Find the area of the largest rectangle having the perimeter of 200 metres.

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288.

Show that among rectangles of given area, the square has the least perimeter.


Let x, y be the sides of rectangle and k2 be the area where k > 0.
therefore space space straight x space straight y space equals space straight k squared space space space space rightwards double arrow space space space space straight y space equals space straight k squared over straight x
Let  straight P space equals space 2 straight x plus 2 straight y space equals space 2 straight x plus 2 straight k squared over straight x                                            open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
       dP over dx equals 0 space space space give space us space 2 minus fraction numerator 2 straight k squared over denominator straight x squared end fraction space equals space 0 comma space space space space space space rightwards double arrow space space space straight x squared space equals space straight k squared space space space space rightwards double arrow space space straight x space space equals straight k comma space space minus straight k
Rejecting x = -k,     we get  x = k
    fraction numerator straight d squared straight P over denominator dx squared end fraction space equals space fraction numerator 4 straight k squared over denominator straight x cubed end fraction
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therefore space space space space straight P space is space minimum space when space straight x space equals space straight k comma space space space space straight y space equals space straight k squared over straight k space equals space straight k space space space space straight i. straight e. space space when space rectangle space is space straight a space square.

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 Multiple Choice QuestionsLong Answer Type

289.

Show that, of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

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 Multiple Choice QuestionsShort Answer Type

290.

Find the dimensions of the rectangle of greatest area that can be inscribed in a semi-circle of radius r.

 
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