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$$\max_{x_1}f(x_1,g(x_1)).$$ And, let $f$ attends max at $x_1^*$, so first order necessary conditions imply that $$\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_1}+\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_2}\dfrac{d g(x_1^*)}{dx_1}=0$$ and as my text says "If the differential function $f(x_1,\dots,x_n)$ reaches a local interior maximum at $(x_1^*,\dots,x_n^*)$, then these hold simultaneously: $$\dfrac{\partial f(x_1^*,\dots,x_n^*)}{\partial x_1}=0;\dots;\dfrac{\partial f(x_1^*,\dots,x_n^*)}{\partial x_n}=0 ",$$ first order necessary conditions also imply that $$\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_1}=0$$ and $$\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_2}=0.$$

Am I right?

Silent
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1 Answers1

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No.

For example, take $f(x,y) = {x^2+y^2 \over 2}$, with $g(x) = 1-x$.

Let $\phi(x) = f(x,g(x)) = (x-{1 \over 2})^2+({1\over 2})^2$.

You can see that the global minimizer of $\phi$ is $\hat{x}={1 \over 2}$, and $g(\hat{x}) = {1 \over 2}$ as well. We have ${\partial \phi(\hat{x}) \over \partial x} = 0$.

However, ${\partial f(\hat{x}, g(\hat{x})) \over \partial x} = \hat{x}$, ${\partial f(\hat{x}, g(\hat{x})) \over \partial y} = \hat{x}$.

One way to think of this is as a constrained minimization $\min \{ f(x,y) | g(x)-y = 0 \}$.

Then at a solution Lagrange multipliers gives $DF((x,y)) + \lambda (Dg(x), -1) = 0$, which simplifies to ${\partial f(\hat{x}, g(\hat{x})) \over \partial x} + {\partial f(\hat{x}, g(\hat{x})) \over \partial y}{\partial g(\hat{x}) \over \partial x} = 0$.

copper.hat
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