Suppose you have the optimization problem

$$ \min_{x,y} f(x,y),\quad (x,y)\in X\subset\mathbb{R}^n\times \mathbb{R}^m \tag{1} $$

Suppose also that the function $f$ is continuous and differentiable, which has a unique stationary point, and this stationary point is also a global minimum.

Since $f$ is diferentiable you can find that $$\frac{\partial f(x,y)}{\partial x}=0$$ gives the necessary condition for the optimal point. Sometimes it gives you one function $ x=g( y)$. But it can gives you more the one implicit functions.

If you have only one function $x=g(y)$, you can substitute into the function and solve the problem $(1)$ as $$ \min_{y} \phi(y),\qquad \phi(y)=f(g( y),{y}),$$ (please see this topic on math.stackexchange).

You can find more related discussions searching for "\(\min_yf(x^*(y),y)\) " on SearchOnMath, like this topic on math.stackexchange.

Note:

1. Theorem. Let $f:X\to \mathbb{R}$ be differentiable on the open convex set $X\subset \mathbb{R}^n\times \mathbb{R}^m$. Then $f$ is quasiconvex on $X$ if and only if $$u,v\in X,\,f(u)\leq f(v)\Longrightarrow \nabla f(v)\cdot(u-v)\leq 0. $$

a) If you can verify the previous theorem then $f$ is a quasi-convex function.

b) If $f$ was a quasi-convex function. It is not clear wy shoud be $\phi(y)=f(g( y),{y})$ convex.

Suppose you have the optimization problem

$$ \min_{x,y} f(x,y),\quad (x,y)\in X\subset\mathbb{R}^n\times \mathbb{R}^m \tag{1} $$

Suppose also that the function $f$ is continuous and differentiable, which has a unique stationary point, and this stationary point is also a global minimum.

Since $f$ is diferentiable you can find that $$\frac{\partial f(x,y)}{\partial x}=0$$ gives the necessary condition for the optimal point. Sometimes it gives you one function $ x=g( y)$. But it can gives you more the one implicit functions.

If you have only one function $x=g(y)$, you can substitute into the function and solve the problem $(1)$ as $$ \min_{y} \phi(y),\qquad \phi(y)=f(g( y),{y}),$$ (please see this topic on math.stackexchange).

You can find more related discussions searching for "\(\min_yf(x^*(y),y)\) " on SearchOnMath, like this topic on math.stackexchange.

Note:

1. Theorem. Let $f:X\to \mathbb{R}$ be differentiable on the open convex set $X\subset \mathbb{R}^n\times \mathbb{R}^m$. Then $f$ is quasiconvex on $X$ if and only if $$u,v\in X,\,f(u)\leq f(v)\Longrightarrow \nabla f(v)\cdot(u-v)\leq 0. $$

a) If you can verify the previous theorem then $f$ is a quasi-convex function.

b) If $f$ was a quasi-convex function. It is not clear why shoud be $\phi(y)=f(g( y),{y})$ convex.

Suppose you have the optimization problem

$$ \min_{x,y} f(x,y),\quad (x,y)\in X\subset\mathbb{R}^n\times \mathbb{R}^m \tag{1} $$

Suppose also that the function $f$ is continuous and differentiable, which has a unique stationary point, and this stationary point is also a global minimum.

Since $f$ is diferentiable you can find that $$\frac{\partial f(x,y)}{\partial x}=0$$ gives the necessary condition for the optimal point. Sometimes it gives you one function $ x=g( y)$. But it can gives you more the one implicit functions.

If you have only one function $x=g(y)$, you can substitute into the function and solve the problem $(1)$ as
  $$ \min_{y} \phi(y),\qquad \phi(y)=f(g( y),{y}). $$

You can find more related discussions searching for "\(\min_yf(x^*(y),y)\) " on SearchOnMath, like this.

Note:

1. Theorem. Let $f:X\to \mathbb{R}$ be differentiable on the open convex set $X\subset \mathbb{R}^n\times \mathbb{R}^m$. Then $f$ is quasiconvex on $X$ if and only if $$u,v\in X,\,f(u)\leq f(v)\Longrightarrow \nabla f(v)\cdot(u-v)\leq 0. $$

a) If you can verify the previous theorem then $f$ is a quasi-convex function.

b) If $f$ was a quasi-convex function. It is not clear wy shoud be $\phi(y)=f(g( y),{y})$ convex.

Suppose you have the optimization problem

$$ \min_{x,y} f(x,y),\quad (x,y)\in X\subset\mathbb{R}^n\times \mathbb{R}^m \tag{1} $$

Suppose also that the function $f$ is continuous and differentiable, which has a unique stationary point, and this stationary point is also a global minimum.

Since $f$ is diferentiable you can find that $$\frac{\partial f(x,y)}{\partial x}=0$$ gives the necessary condition for the optimal point. Sometimes it gives you one function $ x=g( y)$. But it can gives you more the one implicit functions.

If you have only one function $x=g(y)$, you can substitute into the function and solve the problem $(1)$ as $$ \min_{y} \phi(y),\qquad \phi(y)=f(g( y),{y}),$$ (please see this topic on math.stackexchange).

You can find more related discussions searching for "\(\min_yf(x^*(y),y)\) " on SearchOnMath, like this topic on math.stackexchange.

Note:

1. Theorem. Let $f:X\to \mathbb{R}$ be differentiable on the open convex set $X\subset \mathbb{R}^n\times \mathbb{R}^m$. Then $f$ is quasiconvex on $X$ if and only if $$u,v\in X,\,f(u)\leq f(v)\Longrightarrow \nabla f(v)\cdot(u-v)\leq 0. $$

a) If you can verify the previous theorem then $f$ is a quasi-convex function.

b) If $f$ was a quasi-convex function. It is not clear wy shoud be $\phi(y)=f(g( y),{y})$ convex.