0
$\begingroup$

The question should be simple.

Use the Lagrange Multiplier to maximize $f(x,y) = 4x^2 + 10y^2$ subject to the constraint $x^2 + y^2 = 4$. But when I set it up I get two different values for $\lambda$.

$\nabla f(x,y) = \lambda\nabla g(x,y)$

$8x = \lambda2x$ ----> $\lambda = 4$

$20y = \lambda2y$ ---> $\lambda = 10$

$x^2 + y^2 = 4$

What am I missing?

$\endgroup$
  • 2
    $\begingroup$ Missing $0$, maybe because it is so small. The first equation has two solutions, $\lambda=4$ and $x=0$. $\endgroup$ – André Nicolas Apr 9 '14 at 22:38
  • $\begingroup$ See Example $4$ on tutorial.math.lamar.edu/Classes/CalcIII/…. $\endgroup$ – user122283 Apr 9 '14 at 22:49
  • $\begingroup$ Thanks @SanathDevalapurkar, that was exactly what I needed. I wasn't thinking that it was $x = 0$ OR $\lambda = 4$ instead of "and". $\endgroup$ – Brian Barr Apr 10 '14 at 1:28
1
$\begingroup$

When $x=0$ then $y=+2$ or $y=-2$ and $\lambda=10$. When $y=0$ then $x=+2$ or $x=-2$ and $\lambda=4$. The critical points are $(0,2), (0,-2), (2,0), (-2,0)$. By plugging in you can find maxima and minima.

$\endgroup$
  • $\begingroup$ That's what I thought, it seemed fairly straight forward, but I didn't think you were allowed to have multiple values of lambda when using the the Lagrange Multiplier method. $\endgroup$ – Brian Barr Apr 9 '14 at 23:45
0
$\begingroup$

Keeping in mind that what the Lagrange multiplier method seeks out are locations where the "level curves" of the function $ \ f \ $ have normal vectors with the same direction as normal vectors at the intersection points with the constraint function curve, we sometimes have situations where the multiplier itself doesn't have a value that is important or useful.

The two pairs of solution points $ \ ( \ \pm 2 , \ 0 \ ) \ $ and $ \ ( \ 0 , \ \pm 2 \ ) \ $ corresponds to extremal values of the function $ \ f(x,y) \ = \ 4x^2 \ + \ 10y^2 \ $ of 16 and 40, respectively. The geometrical interpretation of this result is the two ellipses that are tangent to the circle $ \ x^2 \ + \ y^2 \ = \ 4 \ $ [in red in the graph below], one interior to the circle ( $ \ 4x^2 \ + \ 10y^2 \ = \ 16 \ $ ) [in blue] , the other outside the circle ( $ \ 4x^2 \ + \ 10y^2 \ = \ 40 \ $ ) [in green] . Since we want the maximal value for the function, we would take the value 40 .

enter image description here

This is likely why we find two values for $ \ \lambda \ $ , one for each extremum. They have no actual application in the solution, however, as is implied in Stefano's answer (and discussed in Paul's Online Notes), since they really tell us nothing about the coordinates of the critical (tangent) points. For $ \ \lambda \ = \ 4 \ $ , we have

$$ 8x \ = \ 4 \ \cdot \ 2x \ , \ 20y \ = \ 4 \ \cdot \ 2y \ \ , $$

which only allows us to infer that $ \ y \ = \ 0 \ $ and nothing about the value for $ \ x \ $ (we get that from the constraint equation). We find similarly, for $ \ \lambda \ = \ 10 \ $ , the other set of tangent points. So the values of $ \ \lambda \ $ only establish the necessity of having either $ \ x \ $ or $ \ y \ $ equal to zero, but it ends up being more important that there are two such "critical values" for the multiplier than what those specific values are.

$\endgroup$

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.