I was looking for example problems online and came across this problem:

a) Use Lagrange multipliers to find the absolute min and absolute max values of $f(x,y)=x^2+4y^2$ subject to the constraint $y=x^2-2$, if they exist

b) Sketch the level set diagram of $f(x,y)=x^2+4y^2$ and the constant curve $y=x^2-2$. Where are the candidate points that the Lagrange multipliers finds?


For a) I get the points $\displaystyle{P=\left(\pm\sqrt{\frac{15}{8}},\frac{-1}{8}\right)}$. I can't real say if they are max or min points the function gives the same value at both points. I know the graph is a paraboloid and was trying to think of the constant curve projected onto the graph. I also tried plugging the constraint into the function and running optimization in one variable. The result is all critical values are complex.

How can we confirm if they are max, min or neither?

Part b) is easy enough to plot and think about level curves being tangent to the constraint curve. Here is a contour plot enter image description here

  • $\begingroup$ Note that $f \to \infty$ as $x \to \pm\infty$. $\endgroup$ – Théophile Oct 13 '15 at 20:33
  • $\begingroup$ ok, then minimum? that would be consistent with the direction the gradient vector say to travel on the constant curve. $\endgroup$ – coffeebelly Oct 13 '15 at 20:36

Everywhere outside the blue ellipse, the value of the function $f$ is larger than the value on the blue ellipse or in its interior. So if you look at $f$'s values on the parabola, they're larger except at the points where it's tangent to that blue ellipse, where there as small as they can be. Hence those two tangent points are minima.

As for maxima, as @Theophile observes, the values of $f$ grow without bound as you move upward on the arms of the parabola.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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