# Constraint Optimization and Lagrange Multipliers (Methods of Optimization)

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.

• Welcome to MSE. I suggest that you tell us which specific problem you are talking about. Oct 12, 2020 at 6:14
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 $$x=\psi(y)$$ $$(a.
Another way of avoiding Lagrange's method is a full parameterization of the constraint curve, after you have understood it. In the case at hand we would take $$\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$$.