# Why does the Lagrange multiplier $\lambda$ change when the equality constraint is scaled?

Consider the problem

$$\begin{array}{ll} \text{maximize} & x^2+y^2 \\ \text{subject to} & \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\end{array}$$

Solving this using the Lagrange multiplier method, I get

$$x = \pm5, \qquad y = 0, \qquad \lambda = 25$$

However, if I rewrite the constraint as $$9x^2+25y^2=225$$, I get a different value of $$\lambda$$, namely, $$\lambda = \frac 19$$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?

• That shouldn't be the case - maybe check for an arithmetic error? – Vasting Apr 7 at 6:34
• Double checked, doesn't seem to be an error. – PGupta Apr 7 at 6:58
• Why wouldn't the multiplier's value change? There's no reason that it shouldn't. – littleO Apr 7 at 7:14
• Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it? – PGupta Apr 7 at 7:17

Let $$f(x,y)=x^2+y^2+\lambda(9x^2+25y^2-225).$$

Thus, from $$\frac{\partial f}{\partial x}=2x+18\lambda x=0$$ and $$\frac{\partial f}{\partial y}=2y+50\lambda y=0$$ we obtain two possibilities: $$\lambda=-\frac{1}{9}$$ or $$\lambda=-\frac{1}{25}.$$

The second gives a minimal value, wile the first gives a maximal value: $$f(x,y)=x^2+y^2-\frac{1}{9}(9x^2+25y^2-225)=25-\frac{16}{9}y^2\leq25,$$ where the equality occurs for $$y=0.$$

If we consider $$f(x,y)=x^2+y^2+\lambda\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right)$$ so we'll get $$\lambda=-25$$

and we'll get the same answer of course.

It happens because $$-\frac{1}{9}\cdot225=-25$$ and $$x^2+y^2-\frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right).$$

• Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225. – PGupta Apr 7 at 7:05
• @PGupta I added something. See now. – Michael Rozenberg Apr 7 at 7:13
• This makes sense, thank you. – PGupta Apr 7 at 7:19

If $$x^\star$$ minimizes $$f(x)$$ subject to the constraint that $$g(x)=0$$, then under mild assumptions there exists a Lagrange multiplier $$\lambda$$ that satisfies $$\tag{1} \nabla f(x^\star) = \lambda \nabla g(x^\star).$$ If $$g$$ is replaced with $$c g$$, then $$x^\star$$ is still a minimizer, but of course $$\lambda$$ no longer satisfies (1). We must correspondingly multiply $$\lambda$$ by $$1/c$$ in order for (1) to remain true.

Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.

• Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this? – PGupta Apr 7 at 7:36
• @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $\delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $c\delta$. The change in the optimal value is the same amount $\epsilon$ in either case. But the rate of change is $\lambda = \epsilon/\delta$ in the first case, and $\lambda/c = \epsilon/(c \delta)$ in the second case. – littleO Apr 7 at 8:05
• Great, got it. Thanks a lot! – PGupta Apr 7 at 9:00

Writing the equation $$\frac{x^2}{25}+\frac{y^2}{9}=1$$ in the form $$y^2=9-\frac{9}{25}x^2$$ you will have the objective function $$f(x)=\frac{16}{25}x^2+9$$