Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I got the function $f(x,y)=e^{-x^2-y^2}$ with the domain of definition $x^2+y^2\leq25$. The task is to decide the biggest and lowest value. How do I get there?

share|cite|improve this question
Do you know about the Second Derivative Test for functions of two variables? – Paul Sundheim Aug 10 '14 at 21:40

$F(x,y)=e^{-x^2-y^2}=e^{-(x^2+y^2)}$. Note that $e^{-z}$ is strictly decreasing with respect to $z$. So to maximize and minimize $F(x,y)$, just minimise and maximise $x^2+y^2$ respectively. By the domain of definition, $x^2+y^2$ is minimised at $0$ and maximised at $25$, so the maximum and minimum values of $F(x,y)$ are $e^{-0}=1$ and $e^{-25}$, respectively.

share|cite|improve this answer
Right. No calculus necessary in this case, since both the function and the domain are radially symmetric, the function in the exponent is decreasing along radii, and the exponential part is an increasing function. – Lubin Aug 10 '14 at 21:48
Don't you just love the exponential function? – Pauly B Aug 10 '14 at 21:53

The minimum or maximum must lie either in the interior region, or on the boundary.

For the interior, you find where the derivative of the function is zero, for both $x$ and $y$.

$\frac{d}{dx}e^{-x^2-y^2} = -2xe^{-x^2-y^2} = 0 $ =>

$x=0$ and similarily $y=0$.

This means we have a possible min or max in $f(0,0) = 1$

Now for the boundary. You can realise the function is constant on all circles with center in (0,0) by inspecting the function closely. $f(x,y)=e^{-x^2-y^2}=e^{-(x^2+y^2)}$, and $x^2+y^2$ is constant on a circle. That means you only have to evaluate one boundary point.

I have chosen $(0,5)$ as my boundary point, and the value of f is $f(0,5)=e^{-25}$.

The minimum and maximum are thus $e^{-25}$ and 1.

An alternative way of solving the boundary would be to parameterise the boundary, and finding the min and max of this function.

Hope this explains it for you.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.