Question

Examine the function `f(x,y)=xy(3-x-y)` for extreme values & find maximum and minimum values of `f(x,y).`

Answer 1

`f(x,y)=xy(3-x-y)=3xy-x^2y-xy^2`

Diff. function w.r.t x and y partially,

`(delf(x,y))/(delx)=3y-2xy-y^2`

`(delf(x,y))/(delx)=0`

`3y-2xy-y^2=0`

`y=0, 3-2x-y=0`

&

`(delf(x,y))/(dely)=3x-x^2-2xy`

`(delf(x,y))/(dely)=0`

`3x=x^2-2xy=0`

`x=0,3-x-2y=0`

Stationary points are : (0,0) ,(3,0) , (0,3) , (1,1)

`r=(del^2f)/(delx^2)=-2y  , t=(del^2f)/(dy^2)=-2x`

`s=(del^2f)/(delxdely)=3-2x-2y`

`s^2=(3-2x-2y)^2`

`rt-s^2=4xy-(3-2x-2y)^2`

For point (0,0), `rt-s^2=-9<0`

The point is of maxima.

For point (3,0), `rt-s^2=-9<0`

The point is of maxima .

For (0,3), `rt-s^2=-9<0`

The point is of maxima.

For point (1,1), `rt-s^2=3>0`

The point is of minima.

(a) Maximum values : At (0,0) , (0,3), (3,0)
At point (0,0) f(max)=0
At point (0,3) f(max)=0
At point (3,0) f(max)=0
(b) Minimum values : At (1,1)
At point (1,1) f(min)=1

The maximum and minimum values of function are 0 and 1.