Finding global maximum of a 2-variable function

calculusglobal optimizationmultivariable-calculus

Let $$f(x,y) = -e^{x^4+y^2}+x^2+y^2+1$$

Find global/local maximum and minimum and upper/lower bound of the function.

I found that the lower bound is ${-\infty} $ and also that, since $$-e^{x^4+y^2}\le -1-x^4-y^2, \forall (x,y)\in \mathbb{R}^2$$we have $$f(x,y)\le -x^4+x^2\le\frac{1}{4},\forall(x,y)\in\mathbb{R}^2$$ so that function is upper bounded.

Then I calculated the critical points, which are $(0,0)$ and $(\pm\alpha,0)$, with $\alpha$ the positive solution of the equation $$e^{x^4}=\frac{1}{2x^2}$$ and by the second-derivative test is apparent that $(0,0)$ is a saddle point while both $(\pm\alpha,0)$ are local maximum.

How do I prove that both $(\pm\alpha,0)$ are global maximum?

Best Answer

The function $f$ is even, so $f(\alpha, 0)=f(-\alpha,0)$. The global maximum of $f$ is in particular a local maximum, hence $(\pm \alpha,0)$ must be the points where $f$ attains a global maximum.

Related Solutions

[Math] Unique critical point does not imply global maximum/global 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:

x^2+y^2(1-x)^3

[Math] Global maximum/ minimum of a function of more than one variable.

A very useful method is Sturm's principle which says that a continuous function on a compact domain (e. g. a closed interval) which is bounded below/above an has at most one local minimum/maximum must have a global minimun/maximum at this very place.

If your function in question, however, has more than one local minimum/maximum, you have to compare the values of the function at all these minima/maxima and those with minimal/maximal value among these candidates are your global minima/maxima.

In each case, you should check first, that your function is bounded below/above!

Thus, for $f:[a,b] \rightarrow \Bbb R$ differentiable, you should proceed as follows:

Calculate the zeros $x_1,x_2,x_3, ...$ of $f'(x)$
Calculate $f(x_1),f(x_2),f(x_3), ...$ as well as f(a)
The minimum of the values calculated in step 2 is your global minimum.

Best Answer

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

[Math] Global maximum/ minimum of a function of more than one variable.

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