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Application of Derivatives

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Short 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.

90 Views

282.

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

96 Views

Long 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?

75 Views

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.

76 Views

Short Answer Type

285.

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

79 Views

286.

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

89 Views

287.

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

98 Views

288.

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

99 Views

Long Answer Type

289.
Show that, of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Let O be the centre of circle of radius a. Let A BCD be the rectangle inscribed in the circle such that AB = x, AD = y.
Now AB² + BC² = AC²∴ x² + y² = 4 a² ...(1)

Let P be the area of rectangle.
$therefore space space space straight P space equals space straight x space straight y therefore space space space space straight P space equals space straight x square root of 4 straight a squared minus straight x squared end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$
$dP over dx space equals straight x. space fraction numerator negative 2 straight x over denominator square root of 4 straight a squared minus straight x squared end root end fraction plus square root of 4 straight a squared minus straight x squared end root space.1 space equals space minus fraction numerator straight x squared over denominator square root of 4 straight a squared minus straight x squared end root end fraction plus fraction numerator square root of 4 straight a squared minus straight x squared end root over denominator 1 end fraction$

$equals fraction numerator negative straight x squared plus 4 straight a squared minus straight x squared over denominator square root of 4 straight a squared minus straight x squared end root end fraction space equals space fraction numerator 4 straight a squared minus 2 straight x squared over denominator square root of 4 straight a squared minus straight x squared end root end fraction$

$dP over dx space equals space 0 space space space gives space us space fraction numerator 4 straight a squared minus 2 straight x squared over denominator square root of 4 straight a squared minus straight x squared end root end fraction space equals space 0 space space space space rightwards double arrow space space space 4 straight a squared minus 2 straight x squared space equals space 0$
$therefore space space space straight x squared space equals space 2 straight a squared space space space space space rightwards double arrow space space space space straight x space equals space square root of 2 space straight a comma space space as space straight x space is space plus ve.$
$fraction numerator straight d squared straight P over denominator dx squared end fraction space equals space fraction numerator square root of 4 straight a squared minus straight x squared end root. left parenthesis negative 4 straight x right parenthesis space minus space left parenthesis 4 straight a squared minus 2 straight x squared right parenthesis. space begin display style fraction numerator negative 2 straight x over denominator 2 square root of 4 straight a squared minus straight x squared end root end fraction end style over denominator square root of 4 straight a squared minus straight x squared end root end fraction$

$equals space fraction numerator begin display style fraction numerator negative 4 straight x square root of 4 straight a squared minus straight x squared end root over denominator 1 end fraction end style plus begin display style fraction numerator straight x left parenthesis 4 straight a squared minus 2 straight x squared right parenthesis over denominator square root of 4 straight a squared minus straight x squared end root end fraction end style over denominator 4 straight a squared minus straight x squared end fraction space equals space fraction numerator negative 4 straight x left parenthesis 4 straight a squared minus straight x squared right parenthesis space plus space straight x left parenthesis 4 straight a squared minus 2 straight x squared right parenthesis over denominator left parenthesis 4 straight a squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction$

$When space straight x space equals space square root of 2 space straight a comma space space fraction numerator straight d squared straight P over denominator dx squared end fraction space equals space fraction numerator negative 4. space square root of 2 straight a space left parenthesis 4 straight a squared minus 2 straight a squared right parenthesis space plus space square root of 2 straight a left parenthesis 4 straight a squared minus 4 straight a squared right parenthesis over denominator left parenthesis 4 straight a squared minus 2 straight a squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction$
$space equals space minus fraction numerator 8 square root of 2 space straight a cubed over denominator space 2 square root of 2 space straight a cubed end fraction space equals space minus 4 space less than space 0$
$therefore space space space straight P space is space maximum space when space straight x space equals space square root of 2 space straight a When space straight x space equals space square root of 2 straight a comma space space straight y space equals space square root of 4 straight a squared minus straight x squared end root space equals space square root of 4 straight a squared minus 2 straight a squared end root space equals space square root of 2. space straight a space space rightwards double arrow space space straight x space space equals straight y space$
$therefore space space space rectangle space becomes space straight a space square.$