Respuesta :
Answer:
The absolute maximum and minimum is [tex]20\; \text{and} -20.[/tex]
Step-by-step explanation:
We first check the critical points on the interior of the domain using the
first derivative test.
[tex]f_x=y-4=0[/tex]
[tex]f_y=x=0[/tex]
The only solution to this system of equations is the point (0, 4), which lies in the domain.
[tex]f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1[/tex]
[tex]\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1<0[/tex]
[tex]\therefore (0,4)[/tex] is a saddle point.
Boundary points - [tex](4,0), (-4,0), (0,16)[/tex]
Along boundary [tex]y=16-x^2[/tex]
[tex]f=x(16-x^2)-4x[/tex]
[tex]=16x-x^3-4x[/tex]
[tex]\Rightarrow f^'=16-3x^2-4=0[/tex]
[tex]\Rightarrow 3x^2=12[/tex]
[tex]\Rightarrow x=\pm2,\;\;y=14[/tex]
Values of f(x) at these points.
[tex]\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}[/tex]
Therefore, the absolute maximum and minimum is [tex]20\; \text{and} -20.[/tex]
