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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?

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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.

Here, $straight s space equals space straight t to the power of 5 space minus space 40 space straight t cubed space minus space 30 space straight t squared space plus space 80 space straight t space minus space 250$
$therefore space space space ds over dt space equals space 5 straight t to the power of 4 minus 120 space straight t squared plus 60 straight t plus 80 space space space space space fraction numerator straight d squared straight s over denominator dt squared end fraction space equals space 20 straight t cubed minus 240 space straight t space plus 60 therefore space space space acceleration space space straight f space equals space 20 space straight t cubed space minus 240 space straight t space plus space 60 space space space space space df over dt space equals 60 space straight t squared minus 240 space space space space space space df over dt space equals space 0 space space space space space space space space space rightwards double arrow space space space 60 straight t squared minus 240 space equals space 0 space space space space space rightwards double arrow space space space space straight t squared minus 4 space equals space 0 space space space space space rightwards double arrow space space space straight t squared space equals space 4 space space space space rightwards double arrow space space space space straight t space space equals negative 2 comma space 2 But space space space straight t not equal to space minus 2 comma space space space space space as space straight t greater than 0 space space therefore space space space space space space straight t space equals space 2 space space space space fraction numerator straight d squared straight f over denominator dt squared end fraction space equals space 120 space straight t At space straight t space space equals 2 comma space space space space space fraction numerator straight d squared straight f over denominator dt squared end fraction space equals space 240 space greater than 0 therefore space space space space space space straight f space is space minimum space at space straight t space equals space 2 therefore space space space space space space minimum space straight f space equals space 20 left parenthesis 2 right parenthesis cubed space minus 240 left parenthesis 2 right parenthesis space plus space 60 space equals space space 160 minus 480 plus 60 equals negative 260 space space space space space space space space space space$