I have two questions regarding the Lagrange Multiplier.
One :
Suppose we are solving this: $$\text{$\max xy$ such that $2x+y=100$}.$$ Solution is: $(x,y)= (25,50)$ and lambda is $25$. So $\text{utility}= 1250$
Suppose we change the constraint: $2x+y=200$. Then, with Lagrange: $(x,y) = (50,100)$ So new $\text{utility=5000}$.
So change in $\text{utility=3750}$
However, if we calculate the change with Lagrange Multiplier, the change in the optimal value should be : $\text{lambda} *\text{change in income} = 25$ into $100= 2500$
Why is this is the case. Shouldn't both methods give the same change in optimal value function. I know that Lagrange Multiplier works for small changes in constraint. But like how small should be the change ? What is the criteria to define small or large change ?
Consider the more general problem:
$$\text{$\max_{x,y} xy$ subject to $2x+y=m$}$$
where $m>0$.
The Lagrangian is
$$L(x,y,\lambda)=xy-\lambda (2x+y-m)$$
A stationary point of the Lagrangian satisfies:
$$\begin{align*} L_x(x,y,\lambda)&=y-2\lambda=0\\ L_y(x,y,\lambda)&=x-\lambda=0\\ L_{\lambda}(x,y,\lambda)&=m-(2x+y)=0\\ \end{align*}$$
The stationary point is
$$(\hat{x},\hat{y},\hat{\lambda})=\left(\frac{m}{4}, \frac{m}{2}, \frac{m}{4}\right).$$
Let $f$ denote the objective function: $f(x,y)=xy$. The maximum of the objective function $\hat{f}$ is given by
$$\hat{f}(m)=f(\hat{x},\hat{y})=f\left(\frac{m}{4}, \frac{m}{2}\right)=\frac{m^2}{8}$$
We have $\hat{f}'(m)=m/4=\hat{\lambda}(m)$ as expected; the Lagrange multiplier represents the marginal change in the maximized objective function with a marginal change in the parameter $m$.
Now consider a change in $m$ from $m_0$ to $m_1$. We have the following linear approximation for the change in $\hat{f}$:
$$\hat{f}(m_1)-\hat{f}(m_0)\approx\hat{f}'(m_0)(m_1-m_0)=\frac{m_0}{4}(m_1-m_0)=\hat{\lambda}(m_0)(m_1-m_0)$$
The exact change in the maximized objective function is
$$\hat{f}(m_1)-\hat{f}(m_0)=\frac{m_1^2-m_0^2}{8}=\frac{(m_0+m_1)}{8} (m_1-m_0)$$
In this example, the exact change can be written as
$$\hat{f}(m_1)-\hat{f}(m_0)=\hat{\lambda}\left(\frac{m_0+m_1}{2}\right)(m_1-m_0).$$
So, if you multiply $\hat{\lambda}$ evaluated at the mean of $m_0+m_1$ (rather than at $m_0$) by the change in $m$, then you will get the exact change in $\hat{f}$.
The fact that you found the linear approximation of the change in the maximized value of the objective function is different from the exact change in the objective function is just a specific illustration of the fact that linear approximations are not exact.
For any differentiable function $g$, the derivative gives only a linear approximation of the change in $g$ for changes in $x$:
$$g(x_1)-g(x_0)\approx g'(x_0)(x_1-x_0).$$
The approximation is very good if $g$ is close to linear or $x_1-x_0$ is small, but otherwise may not be a good approximation.