Global max/min of surfaces Given $f(x,y)=4x^3+4x^2y+3y^2$ and restrictions $x,y≥0$ and $x+y≤1$, I'm trying to find global max and mins.
I found the partial derivative and found the critical point $(0,0)$ by setting those to $0$, but I'm not sure how to get other critical points. I know it has something to do with taking the traces, but I'm unsure how to go about doing this. 
 A: From your restrictions you obtain $x \leq 1 - y$ and because $y \geq 0$ you will only have $x \geq 0$ if $0 \leq y \leq 1$. Also it follows $0 \leq x \leq 1$. Now taking partial derivatives of $f(x,y)$ one obtains the equations:
1) $12x^2+8xy=0$
2) $4x^2+6y=0$
1) has Solutions $a)x=0$ and $b)x = \frac{-2y}{3}$. But solution b) cannot be a solution under above restrictions (exception: $y=0$; in this case it would also follow $x=0$).
2) has Solution $y=- \frac{2x^2}{3}$; the only possible value that $x$ can take is $x=0$ because otherwise $y$ would be negative (excluded by the restrictions).
Therefore the only critical Point which respects above restrictions is $(0,0)$.
The Maximum will occur on the endpoints of the interval of $x$ and $y$ since $f(x,y)$ is increasing with both $x$ and $y$. The Minimum will occur at $x=0,y=0$ where the function $f(x,y)$ has the smallest value under These restrictions.
A: $(0,0)$ clearly gives the minimum, but the maximum of the function is not at a critical point. Since $3x^2 + 3y^2$ is a convex function, your maximum will occur at the point $(1,0)$ given your restrictions on $x,y$.
A: $p,q \ge0,p+q=1,\alpha=\dfrac{x+y}{p+q} \le 1  ,x=\alpha p, y=\alpha q \implies f(x,y)=4\alpha^3p^3+4\alpha^3p^2q+3\alpha^2q^2 \le 4p^3+4p^2q+3q^2=4p^2+3(1-p)^2=7p^2-6p+3=g(p)$
it is trivial $g(p)$ has max point on $p=0$ or $p=1$ and it is easy to verify $g_{max}=g(1)=4$
