Quote:

Originally Posted by

**jacob93** $\displaystyle f(x,y)=x+xy^2$

only def for $\displaystyle x^2+y^2 \leq 4$

find the highest and lowest value for $\displaystyle f$.

Consider

$\displaystyle f(x,y)=x+xy^2 = x(1+y^2)$ $\displaystyle f(x,y) \ge 0 x \ge 0$ and $\displaystyle f(x,y) < 0 x < 0$ one conclusion that can be drawn is $\displaystyle f(x,y)$ reaches its extremes under the restriction $\displaystyle x^2+y^2 \leq 4$ at precisely on the circle $\displaystyle x^2+y^2 = 4$. So substitute $\displaystyle y^2 = 4 -x^2 $ in $\displaystyle f(x,y)$. We have $\displaystyle g(x) = x(5-x^2)$ Equating the derivative to zero we have $\displaystyle g'(x) = 5- 3x^2 = 0 \implies x = \pm \sqrt{\frac{5}{3}}$