# In Lagrange multiplier method, what does a larger $\lambda$ multiplier mean?

In the Lagrange multiplier method, let us say we have a constraint $f(x) \le C$.

So we say setup the lagrange $L = ... -\lambda (C - f(x))$ where "..." contains the objective function.

My question is, let us say $\lambda$ in our optimal solution is really large. What does that tell us about the constraint?

• – Winther Jun 5 '18 at 19:51

If the constraint is given by, say, $$x^2+y^2-1=0$$ you will get one value of $\lambda$. If you instead use $$2x^2+2y^2-2=0$$ then you get a different value for $\lambda$. Continuing this way, you can make any non-zero value for $\lambda$ you want from the exact same constraint, just by expressing the constraint differently (as long as there is a non-zero value for $\lambda$ in the first place).
• But if you fix the constraint equation then wouldn't different values of $\lambda$ at different points tell you something about the sensitivity of the objective function to changes in input? – Mark S. Jun 5 '18 at 19:52
• It depends on the context and the problem. As here where OP has defined the constraint as $\lambda (C - f(x))$ where $C$ seems to be some fixed parameter then it has a meaning through $\lambda = \frac{dL}{dC}$. – Winther Jun 5 '18 at 19:56
• If you have a fixed expression for the constraint, sure, the value of $\lambda$ tells you something. But it doesn't tell you anything directly about the function's actual behaviour because it is, to such a large degree, an artifact of our $f$. Also, comparing $\lambda$'s at different points might give us some info, but again it could just be an artifact of our chosen $f$ (compare my original example to something like $(x^2+y^2-1)(x^2+(y-1)^2+1)=0$, which is still the same circle, but where the gradient of the left-hand side does not have the same length at all points). – Arthur Jun 5 '18 at 20:33
In the case of Dido's problem, you want to maximalize the integral $$I[y]=\int_{x_1}^{x_2}y \mathrm{d}x$$ with the condition $$J[y]=\int_{x_1}^{x_2}\sqrt{1+y'^2} \mathrm{d}x$$ So your Lagrange function will be $$L=y+\lambda \sqrt{1+y'^2}$$ To this problem, the solution will be a circle, with radius $|\lambda|$. So it can have a meaning.