Question

find maximum and minimum value of xy(a-x-y)

Answer 1

it is a stupid question

Answer 2

Answer:

we have local maximum at ( 1/3 , 1/3 )

Step-by-step explanation:

F ( x ) = xy ( a-x-y )

F ( x ) = axy - x²y - xy²

Taking partial derivative wrt x

F ' ( x ) = ay - 2xy - y²

Taking partial derivative wrt y

F ' ( y ) = ax - x² - 2xy

Taking second partial derivative wrt x

F '' ( x ) = - 2y

Taking second partial derivative wrt y

F '' ( y ) = - 2x

Also,

F' ( xy ) = a - 2x - 2y

Now,

F ' ( x ) = 0

ay - 2xy - y² = 0

y ( a - 2x - y ) = 0

y = 0 OR  a - 2x - y = 0

y = 0 , y = a - 2x

Now taking F ' ( y ) = 0

F ' ( y ) = ax - x² - 2xy

Taking y = 0 [ as we got the value of y = 0 and a - 2x ]

0 = ax - x² - 2x ( 0 )

0 = x ( a - x )

x = 0 or x = a

The points are ( 0 , 0 ) and ( a , 0 )

Taking y = a - 2x

0 = ax - x² - 2x ( a - 2x )

0 = ax - x² - 2ax + 4x²

0 = 3x² - ax

0 = x ( 3x - a )

x = 0 or x = a/3

Substituting x in y = a - 2x

So now the coordinates are ( 0 , a ) and ( a/3 , a/3 )

We know,

D = F '' ( x ) F'' ( y ) - [ F' ( x y) ] ²

D = ( - 2y ) ( - 2x ) - [ 1 - 2x - 2y ] ²

Now evaluating D on the four coordinates :

D ( 0, 0 ) = - 1

D ( a , 0 ) = - 1

D ( 0 , a ) = - 1

Taking F '' ( x ) on ( 1/3, 1/3)

= -2/3 < 0

Therefore we have local maximum at ( 1/3 , 1/3 )