## Calculus: Early Transcendentals 8th Edition

Absolute maximum: $f(1,\sqrt 2)=2$ Absolute minimum: $f(0,0)=0$ (as well as other points along $(0,y)$ and $(x,0)$)
To calculate the critical points we have to first put $f_x(x,y), f_y(x,y)$ equal to $0$. Thus, $f_x=y^2,f_y=2xy$ This yields, $x=0,y=0$ and $f(0,0)=0$ From the the given boundary $x^2+y^2=3$, we have $y=\sqrt {3-x^2}$ $f(x)=x(3-x^2)$ and $f'(x)=3-3x^2$ so, $x=1$ and $y=\sqrt {3-x^2} =\sqrt 2$ Thus, Critical point: $(1,\sqrt 2)$ At $(x,y)=(1,\sqrt 2)$ $f(1,\sqrt 2)=2$ Therefore, it has minimum and maximum values at $0,2$ Hence, Absolute maximum: $f(1,\sqrt 2)=2$ Absolute minimum: $f(0,0)=0$ (as well as other points along $(0,y)$ and $(x,0)$)