maxima-minimamultivariable-calculuspolynomials
Given the function
$$f(x,y)={x^3-3xy+y^3-2}.$$ defined in the rectangle $[0,3] \times [0,2]$. Find the global minimum and maximum and at what point does it reach.
I tried finding the maximum, it is $25$ and $(3,0)$, but I am not sure. I haven't tried the minimum.
Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.
A picture is tremendously helpful. If you have access to maple, try this:
plots[animate](plot,
[x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);
Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:
This function is convex, all local minimum is a global minimum. $(0,0)$ is the global minimum.
We can also prove it by completing the squares.
\begin{align}Ax^2+2Bxy+Cy^2 +D &= A \left(x+\frac{By}{A} \right)^2+Cy^2-A\left(\frac{By}{A} \right)^2+D \\ &=A \left(x+\frac{By}{A} \right)^2+\left(C-\frac{B^2}{A} \right)y^2+D \\&=A \left(x+\frac{By}{A} \right)^2+\left(\frac{AC-B^2}{A} \right)y^2+D \\ &\ge D\end{align}
Hence $D$ is the lower bound which can only be attained when $y=0$ and $x=0$.
Best Answer
Let $f(x,y)=x^3-3xy+3y^3-2$.
Thus, $f$ is a convex function of $x$ and $f$ is a convex function of $y$.
Since a convex function gets a maximal value for an extreme value of the variable, we obtain: $$\max_{(x,y)\in[0,3]\times[0,2]}f=\max_{x\in\{0,3\},y\in\{0,2\}}f=f(3,0)=25.$$ Now, $f(1,1)=-3$ and by AM-GM: $$x^3-3xy+y^3-2-(-3)=x^3+y^3+1-3xy\geq3\sqrt[3]{x^3y^3}-3xy=0,$$ which says that $-3$ is a minimal value.