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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.

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282.

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

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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?

Let x and y be two numbers.
$therefore space space space straight x plus straight y space equals space 1 half space space space or space space space space straight y space equals space 1 half minus straight x$ ...(1)
Let $straight S space equals space straight x squared plus straight y cubed space equals straight x squared plus open parentheses 1 half minus straight x close parentheses cubed space equals space straight x squared plus 1 over 8 left parenthesis 1 minus 2 straight x right parenthesis cubed$ $open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$
$equals straight x squared plus 1 over 8 left parenthesis 1 minus 6 straight x plus 12 straight x squared minus 8 straight x cubed right parenthesis space equals space minus straight x cubed plus 5 over 2 straight x squared minus 3 over 4 straight x plus 1 over 8$

$dS over dx space equals space minus 3 straight x squared plus 5 straight x minus 3 over 4$

$dS over dx space equals space 0 space space space space space rightwards double arrow space minus 3 straight x squared plus 5 straight x minus 3 over 4 space equals space 0 space space space space rightwards double arrow space space space space 12 straight x squared minus 20 straight x plus 3 space equals space 0$

$therefore space space space space straight x space equals space fraction numerator 20 space plus-or-minus square root of 400 minus 144 end root over denominator 24 end fraction space equals space fraction numerator 20 plus-or-minus square root of 256 over denominator 24 end fraction space equals space fraction numerator 20 plus-or-minus 16 over denominator 24 end fraction equals space space fraction numerator 20 plus-or-minus 16 over denominator 24 end fraction equals space 3 over 2 comma space 1 over 6$

$fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space minus 6 straight x plus 5$
When $straight x equals space 3 over 2 comma space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space minus 9 plus 5 space equals space minus 4 less than 0$

$therefore space space space space straight S space is space maximum space when space straight x space equals space 3 over 2$

$When space straight x space equals space 1 over 6. space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space minus 1 plus 5 space equals space 4 greater than 0 therefore space space space space straight S space is space minimum space when space straight x space equals space 1 over 6 When space straight x space equals space 1 over 6 comma space space space straight y space equals space 1 half minus 1 over 6 space equals space fraction numerator 3 minus 1 over denominator 6 end fraction space equals space 2 over 6 equals space 1 third space space space space space space space space space space space space space space space space space space space space left square bracket because space space space of space space left parenthesis 1 right parenthesis right square bracket therefore space numbers space are space 1 over 6 comma space space 1 third.$

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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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Short 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.

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Long 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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Short 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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Short Answer Type

Long Answer Type

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

Short Answer Type

Long Answer Type

Short 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?