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Application Of Derivatives

Question
CBSEENMA12035501

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

Solution

Let the two parts be x,   15-x.
Let       straight y space equals space straight x squared left parenthesis 15 minus straight x right parenthesis cubed
   dy over dx space equals space straight x squared. space straight d over dx left square bracket left parenthesis 15 minus straight x right parenthesis cubed right square bracket space plus space left parenthesis 15 minus straight x right parenthesis cubed. space straight d over dx left parenthesis straight x squared right parenthesis
space space space space space space space space space equals space straight x squared.3 space left parenthesis 15 minus straight x right parenthesis squared. space left parenthesis negative 1 right parenthesis space plus space left parenthesis 15 minus straight x right parenthesis cubed. space 2 straight x
space space space space space space space space space space equals space straight x left parenthesis 15 minus straight x right parenthesis squared. space left square bracket negative 3 straight x plus 2 left parenthesis 15 minus straight x right parenthesis right square bracket space equals space straight x left parenthesis 15 minus straight x right parenthesis squared left parenthesis negative 5 straight x plus 30 right parenthesis
space space space space dy over dx space equals space 0 space give space us space straight x left parenthesis 15 minus straight x right parenthesis squared space left parenthesis negative 5 straight x plus 30 right parenthesis space equals space 0 space space space rightwards double arrow space space space space straight x space equals space 0 comma space space 15 comma space space 6.
Rejecting x = 0,  15 as far these values of x, we have y = 0, which is not possible. 
therefore space space space space we space have space straight x space equals space 6
fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space straight x left parenthesis 15 minus straight x right parenthesis squared. space left parenthesis negative 5 right parenthesis space plus space straight x space left parenthesis negative 5 straight x plus 30 right parenthesis. space space 2 left parenthesis 15 minus straight x right parenthesis space left parenthesis negative 1 right parenthesis space plus space left parenthesis 15 minus straight x right parenthesis squared thin space left parenthesis negative 5 straight x plus 30 right parenthesis. space 1
At space space straight x space equals space 6. space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals 6 space left parenthesis 15 space minus 6 right parenthesis squared. space left parenthesis negative 5 right parenthesis space plus space 0 space plus space 0 space equals space minus 2430 space less than 0
therefore space space space space straight y space is space maximum space value space when space straight x space equals space 6 space space space space space rightwards double arrow space space space space two space parts space are space 6 comma space 15 space minus 6 space straight i. straight e. comma space 6 comma space 9.

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