Question

Find the absolute extrema of the function f(x, y) = x^2 + 3y^2 + 12y on the closed, bounded region x^2 + y^2 ≤ 9

Answer 1

Answer:

The function ,f(x,y) is    

       $=x^ 2+3 y^2 +12 y$

Partial Derivative of f(x,y) with respect to x and y is :

         $\frac{\partial f(x,y)}{\partial x }=2 x=0\\\\ \frac{\partial f(x,y)}{\partial y }=6 y +12$

Critical points are , x=0 and , 6 y=- 12, y= -2, that is (0,-2).

The Closed bounded region in which we have to find extrema is,

   $x^2 + y^2 \leq   9$

which is the equation of the circle, having center (0,0) radius equal to 3 units.

The critical point (0,-2), lies inside the bounded region.the other critical points will be (3,0),(-3,0), (0,3) and (0,-3).

We will use second derivative test to find the Extrema of the function

Second Derivative Test

If  f(x, y) has continuous second-order partial derivatives in some open disk containing the point (a, b) and that fx (a, b) = fy (a, b) = 0. The discriminant M(a, b) is given  by

$M(a, b) = f_{xx} (a, b)f_{yy}(a, b) - [f_{xy} (a, b)]2$

1.

$f_{xx}=2x *2 x=4 x^2, f_{yy}=(6 y+12)*6=36*(y+2)$

M(0,-2)= 0× 36×(-2+2) - [0²+3 (-2)²+12×(-2)]²

= 0 - (12 -24)²

= - 144

also, $f_{xx}(0,-2)=0$

, f(0,-2)=12-24=-12

2.

M(3,0)=36 * 72 - (9)=2592-9=2583>0

$f_{xx}(3,0)=4*3^2=36$

f(3,0)=9

3. M(-3,0)=36 * 72 - (9)=2592-9=2583>0

$f_{xx}(-3,0)=4*3^2=36$

f(-3,0)=9

4. M(0,3)= 0- 27-36=-63<0

$f_{xx}(0,3)=27+36=63$

5. M(0,-3)=0 - 27+36=9>0

$f_{xx}(0,-3)=4*0^2=0$

f(0,-3)=27-36=-9

Point of maxima or extreme = (0,3)=63

Point of minima =(0,-2)=-12