Answer:
At critical point in D
a
   [tex](x,y) = (0,0)[/tex]
b
[tex]f(x,y) = f(x) =11 -x^2[/tex]
where [tex]-1 \le x \le 1[/tex]
c
maximum value 11
minimum value  10
Step-by-step explanation:
Given [tex]f(x,y) =10x^2 + 11x^2[/tex]
At critical point
[tex]f'(x,y) = 0[/tex]
 =>  [tex][f'(x,y)]_x = 20x =0[/tex]
=> Â [tex]x =0[/tex]
Also
[tex][f'(x,y)]_y = 22y =0[/tex]
=> Â [tex]y =0[/tex]
Now considering along the boundary
    [tex]D = 1[/tex]
=> Â [tex]x^2 +y^2 = 1[/tex]
=> Â [tex]y =\pm \sqrt{1- x^2}[/tex]
Restricting [tex]f(x,y)[/tex] to this boundary
   [tex]f(x,y) = f(x) = 10x^2 +11(1-x^2)^{\frac{2}{1} *\frac{1}{2} }[/tex]
              [tex]= 11-x^2[/tex]
At boundary point D = 1
Which implies that [tex]x \le 1[/tex] Â or [tex]x \ge -1[/tex]
So the range of  x is
         [tex]-1 \le x \le 1[/tex]
Now along this this boundary the critical point is at
      [tex]f'(x) = 0[/tex]
=> Â Â Â Â [tex]f'(x) = -2x =0[/tex]
=> Â Â Â Â [tex]x=0[/tex]
Now at maximum point [tex](i.e \ x =0)[/tex]
      [tex]f(0) =11 -(0)[/tex]
          [tex]= 11[/tex]
For the minimum point x = -1 or x =1
       [tex]f(1) = 11 - 1^2[/tex]
           [tex]=10[/tex]
       [tex]f(-1) = 11 -(-1)^2[/tex]
             [tex]=10[/tex]
     Â
      Â
