Mathematical explanation of the intuition behind the Lagrange multiplier Suppose we are to maximize $U_I(x,y)$ subject to $p_xx+p_yy = I$ where $U_I(x,y)$ is differentiable everywhere and $\frac{\partial U}{\partial x} > 0$, $\frac{\partial U}{\partial y} > 0$ and $U$ is quasi-concave. The latter properties of $U$ are given to ensure that the Kuhn-tucker conditions hold.
If $I' > I$, can we then say that $U_{I'}(x^{*},y^{*}) - U_{I}(x^{*},y^{*}) = (I'-I)\lambda$?
I read that when there's a change in the constraint from $p_xx+p_yy = I$ to $p_xx + p_yy = I'$ (where $I' > I$), the change in $U^{*}$ will be given by $(\Delta I) \cdot \lambda$. I verified that this works for the Cobb-Douglas production function $U(x,y) = x^a y^b$, but does it work in general? In other words, is it true that $$\frac{\Delta U_{optimal}}{\Delta I} = \lambda?$$
Update: This might be useful although I don't know how exactly.
 A: From the f.o.c. for utility maximization under a budget constraint, we have
$$\frac{\partial U}{\partial x} = \lambda p_x,\;\;\;\frac{\partial U}{\partial y} = \lambda p_y$$
that holds at the optimal quantities $(x^*, y^*)$.
Set $V = U(x^*, y^*)$ and then (assuming fixed prices)
$$\frac{dV}{d I} = \frac{\partial U}{\partial x}\cdot \frac{d x^*}{d I} + \frac{\partial U}{\partial y}\cdot \frac{d y^*}{d I}.$$
Using the f.o.c we get
$$\frac{d V}{d I} = \lambda p_x\cdot \frac{d x^*}{d I} + \lambda p_y\cdot \frac{dy^*}{d I}$$
$$\implies \frac{d V}{d I} = \lambda \cdot \left(p_x\cdot \frac{d x^*}{d I} + p_y\cdot \frac{d y^*}{d I}\right).$$
Multiply throughout by $dI$ to get
$$\implies d V = \lambda \cdot \left(p_x\cdot d x^* + p_y\cdot d y^*\right).$$
But the term in parentheses is the change in income, $dI$, so
$$d V = \lambda \cdot dI.$$
So this holds for infinitesimal changes in income, and it can be an acceptable approximation to discrete but "small" changes in $I$.
Note: In general, multipliers on the constraints represent the marginal change in the value of the optimized objective function when the constraint is "relaxed" (here, an increase in income).
