Respuesta :
Answer:
absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.
Step-by-step explanation:
We compute,
[tex]$ f_x = 6x^2, f_y=4y^3 $[/tex]
Hence, [tex]$ f_x = f_y = 0 $[/tex] if and only if (x,y) = (0,0)
This is unique critical point of D. The boundary equation is given by
[tex]$ x^2+y^2=1$[/tex]
Hence, the top half of the boundary is,
[tex]$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}[/tex]
On T we have, [tex]$ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1 \text{ for}\ -1 \leq x \leq 1$[/tex]
We compute
[tex]$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$[/tex]
0 if and only if x=0, x= 1/2 or x = -2.
We disregard [tex]$ x = -2 \notin [-1,1]$[/tex]
Hence, the critical points on T are (0,1) and [tex]$(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$[/tex]
On the bottom half, B, we have
[tex]$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$[/tex]
Therefore, the critical points on B are (0,-1) and [tex]$( 1/2, -\sqrt3/2)[/tex]
It remains to evaluate f(x, y) at the points [tex]$ (0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\ (\pm1, 0)$[/tex] .
We should consider latter two points, [tex]$(\pm1,0)$[/tex], since they are the boundary points for the T and also B. We compute [tex]$ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $[/tex]
We conclude that the absolute maximum = f(1, 0) = 2
And the absolute minimum = f(−1, 0) = −2.
