Respuesta :
Solution :
a). Let [tex]$f(x)=5ye^x-e^{5x}-y^5$[/tex]
[tex]$f_x=0 \Rightarrow 5ye^x-5e^{5x}=0$[/tex]
[tex]$\Rightarrow 5e^x[y-e^{4x}]=0$[/tex]
[tex]$f_y=0 \Rightarrow 5e^x-5y^4=0$[/tex]
[tex]$\Rightarrow 5(e^x-y^4)=0$[/tex]
[tex]$y=e^{4x}$[/tex] and [tex]$y^4=e^x$[/tex] [tex]$\Rightarrow y=(y^4)^4=0$[/tex]
[tex]$y^{16}-y=0 \Rightarrow y[y^{15}-1]=0$[/tex]
Therefore, y = 0 and y = 1
If y = 0, [tex]$e^{4x}=0 $[/tex] (not defined)
If y = 1, [tex]$e^{4x}=1 \Rightarrow e^{4x} = \ln (1) \Rightarrow x=0$[/tex]
∴ (0,1) is the only critical point.
[tex]$f_{xx}= 5ye^x-25e^{5x}$[/tex]
[tex]$f_{xy} = 5e^x, f_{yy} = -20y^3$[/tex]
At (0,1), [tex]$f_{xx}=-20, f_{xy} = 5, f_{yy} = -20$[/tex]
[tex]$f_{xx}f_{yy}-f^2_{xy}=400-25(>0)$[/tex]
∴ f has local maximum at (0,1)
b). On y axis, x = 0
[tex]$f(y)= 5y-1-y^5$[/tex]
[tex]$f_y=5-5y^4=0$[/tex]
[tex]$y^4=1 \Rightarrow y = 1,-1 \notin \text{domain}$[/tex]
[tex]$f_{yy} = -20y^3$[/tex]
At y = 1, [tex]$f_{yy} = -20<0$[/tex]
∴ f has local maximum.
At y = -1, [tex]$f_{yy} = 20>0$[/tex]
f has local minimum
Since y extends infinitely, f has no global maximum.
