Newbie question here. So I am starting to learn about constrained optimization in my multivariable calculus course and I was taught how to use the Lagrangian and Lagrange multipliers to solve an optimization problem subject to a specific constraint. I was also taught before this how to solve an optimization problem without using the Lagrangian by converting the objective function into a single variable one using the constraint equation and finding its critical point.

Now, when I did a problem subject to an equality constraint using the Lagrange multipliers, I succeeded to find the extrema. However, when I did the method where you restrict the objective function into a single variable using the constraint equation, there is no critical point and hence I could not solve. What seems to be the problem here?

This is the specific problem:

Let f(x,y) = (x^2) + (y^2) - 2y. Find the absolute extrema on the circle: (x^2) + (y^2) = 4.

Using the Lagrange multipliers, I found the absolute maximum at (0,-2) and absolute minimum at (0,2).

If I, instead, substitute the (x^2) term on my objective function as (4 - (y^2)), it becomes 4 - 2y. Getting the derivative with respect to y gives -2 and hence it has no critical points.

  • $\begingroup$ Welcome to MSE. I suggest that you tell us which specific problem you are talking about. $\endgroup$ Oct 12, 2020 at 6:14
  • $\begingroup$ I added the problem $\endgroup$
    – Lyle
    Oct 12, 2020 at 6:22

1 Answer 1


A constraint $g(x,y)=0$ can define a complicated geometric object, e.g., a lemniscate. You cannot expect that a single variable $y$ will be able to model this object. In the case at hand we have the constraint $g(x,y):=x^2+y^2-4$ that defines a circle $C$. Furthermore $\nabla g(x,y)=(2x,2y)\ne(0,0)$ at all points of $C$. This guarantees that Lagrange's method will work.

Now your elimination of the variable $x$ has left you with the new objective function $\hat f(y):=4-2y$. It is true that at all points $(x,y)\in C$ we have $f(x,y)=\hat f(y)=4-2y$, and $f\restriction C$ is extremal where $\hat f\restriction C$ is extremal, i.e. at the points $(x,y)\in C$ where $y$ is maximal or minimal. These are the points $(0,\pm2)$ which you ad already found using Lagrange's method. But "elimination of a variable" is an obscure algebraic process which is in many cases not apt to master the complete geometric situation. In fact you cannot describe the full circle $C$ as a graph of some auxiliary function $x=\psi(y)$ $(a<y<b)$.

Another way of avoiding Lagrange's method is a full parameterization of the constraint curve, after you have understood it. In the case at hand we would take $$\phi\mapsto(x(\phi),y(\phi)\bigr):=(2\cos\phi,2\sin\phi)\qquad\bigl(\phi\in{\mathbb R}/(2\pi)\bigr)$$ and obtain the pullback $$\hat f(\phi):=f\bigl(x(\phi),y(\phi)\bigr)=4-4\sin\phi\ .$$ It is now easy to find the extrema of this pullback and the corresponding points $(x,y)\in C$.


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