Question

From Hardcover Book, Marsden/Tromba, Vector Calculus

Answer 1

Solution :

a). Let $f(x)=5ye^x-e^{5x}-y^5$

$f_x=0 \Rightarrow 5ye^x-5e^{5x}=0$

         $\Rightarrow 5e^x[y-e^{4x}]=0$

$f_y=0 \Rightarrow 5e^x-5y^4=0$

         $\Rightarrow 5(e^x-y^4)=0$

$y=e^{4x}$ and $y^4=e^x$ $\Rightarrow y=(y^4)^4=0$

$y^{16}-y=0 \Rightarrow y[y^{15}-1]=0$

Therefore, y = 0 and y = 1

If y = 0, $e^{4x}=0$ (not defined)

If y = 1, $e^{4x}=1 \Rightarrow e^{4x} = \ln (1) \Rightarrow x=0$

∴ (0,1) is the only critical point.

$f_{xx}= 5ye^x-25e^{5x}$

$f_{xy} = 5e^x, f_{yy} = -20y^3$

At (0,1), $f_{xx}=-20, f_{xy} = 5, f_{yy} = -20$

$f_{xx}f_{yy}-f^2_{xy}=400-25(>0)$

∴ f has local maximum at (0,1)

b). On y axis, x = 0

  $f(y)= 5y-1-y^5$

  $f_y=5-5y^4=0$

  $y^4=1 \Rightarrow y = 1,-1 \notin \text{domain}$

$f_{yy} = -20y^3$

At y = 1, $f_{yy} = -20$

∴ f has local maximum.

At y = -1, $f_{yy} = 20>0$

f has local minimum

Since y extends infinitely, f has no global maximum.