Firstly plot the graphs of the given functions that define the region D and find the points of intersection You should be finding the
points A( -3, 0 ) and B( 4,7)
.
Then let's find the critical points in the interior of region D and the values of the function at these points
fx = 2xy and fy = x2. The only critical point in D is the (0,0) and f(0,0) = 0.
Now let's turn our attention at the boundary of the region D.
The parabola y = x2 -9 and the line y= x +3 , both on the interval [0,4) comprise the boundary.
- If y = x2 -9 then f(x,y) = f(x) = x2(x2-9) = x4−9x2. then f' (x) = 4x3−18x = 2x(2x2−9)
which has x =0, x= 3/√2, and x=- 3/√2 as critical point in the interval of interest [0,4].
The values of f at these points are 0, -81/4.
- If y= x+3 then f(x,y) = x2(x+3)= x3 +3x2 . Then f'(x) = 3x2+6x and its critical points
x=0 and x= -2. Where the function f takes values 0 and 4.
Finally let's see the values that the function takes at the points of intersection A( -3, 0 ) and B( 4,7)
At A the function assumes the value of 0 and at B the value of 112.
If we sort the values of f we see that the function in the region D assumes its global minimum of -81/4 and its global or absolute maximum of 112 at (4,7)