Sponsor Area

Application Of Derivatives

Question
CBSEENMA12035504

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?

Solution

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.