I want to find the extrema of a function $F(x,y,z)$ subject to a constraint $g(x,y,z)=c$. This means that implicitely $z=z(x,y)$. So I find $ \frac {\partial F}{\partial x}$ and $ \frac {\partial F}{\partial y}$, using the chain rule to take into account that $z=z(x,y)$. Then I find the points $(x,y,z)$ that solve $\frac {\partial F}{\partial x}=0$ and $\frac {\partial F}{\partial x}=0$.

For example, let $F(x,y,z)=20+2x+2y+z^2$ and $g(x,y,z)=x^2+y^2+z^2=11$. So $$\frac {\partial F}{\partial x}=2+2z\frac {\partial z}{\partial x}$$ and $$\frac {\partial F}{\partial y}=2+2z\frac {\partial z}{\partial y}$$ To find $\frac {\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ I use the constraint equation, taking into consideration that the implicit derivatives are $\frac{\partial z}{\partial x}=-\frac{g_x}{g_z}$ and $\frac{\partial z}{\partial y}=-\frac{g_y}{g_z}$. Given that $g_x=2x$, $g_y=2z$ and $g_z=2z$, then $\frac{\partial z}{\partial x}=-\frac{2x}{2z}=-\frac{x}{z}$ and $\frac{\partial z}{\partial y}=-\frac{2y}{2z}=-\frac{y}{z}$. So, $$0=\frac {\partial F}{\partial x}=2+2z\frac {-x}{z}=2(1-x)$$ and $$0=\frac {\partial F}{\partial y}=2+2z\frac {-y}{z}=2(1-y)$$ We see that $x=1$ and $y=1$. Replazing this values in the constraint equation we found $z$. $$1^2+1^2+z^2=11 \Rightarrow z=\pm 3$$ We can check that $(x,y,z)=(1,1,3)$ and $(x,y,z)=(1,1,-3)$ are maxima.

The real question start here. How to apply this method of optimization (I think it's called "constrained optimization with implicit differentiation") if I have two (o more) constraint equations instead of one?

For example, I want to optimize the (same) function $F(x,y,z)=20+2x+2y+z^2$ constrained by $g_1(x,y,z)=x^2+y^2+z^2=11$ and $g_2(x,y,z)=x+y+z=3$.

$\frac {\partial F}{\partial x}$ and $\frac {\partial F}{\partial y}$ still are $$\frac {\partial F}{\partial x}=2+2z\frac {\partial z}{\partial x} \tag{1} $$ and $$\frac {\partial F}{\partial y}=2+2z\frac {\partial z}{\partial y} \tag{2} $$ but now $$\frac{\partial z}{\partial x}=-\frac{{g_1}_x}{{g_1}_z}=-\frac{2x}{2z}=-\frac{x}{z} \ \ and \ also \ \ \frac{\partial z}{\partial x}=-\frac{{g_2}_x}{{g_2}_z}=-\frac{1}{1}=-1 \tag{3}$$ and $$\frac{\partial z}{\partial y}=-\frac{{g_1}_y}{{g_1}_z}=-\frac{2y}{2z}=-\frac{y}{z} \ \ and \ also \ \ \frac{\partial z}{\partial y}=-\frac{{g_2}_y}{{g_2}_z}=-\frac{1}{1}=-1 \tag{4}$$ (This might suggest that the extrema satisfies $x=y=z$, but I've checked with other method of optimiztion and this is wrong.) So, which of this $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ should I plug into equations $(1)$ and $(2)$ in order to find the extrema of $F(x,y,z)$ constrained?

(I know that I could solve this problem using direct substitution or Lagrange multipliers, but that's not what I ask in this question)

Edit: Reading the answers and reading about the implicit function theorem (in chapter 2 of Advanced Calculus, by Kaplan), finally I understood how to solve it.

For handier notation I rename the constraints $g_1(x,y,z)=c_1$ and $g_2(x,y,z)=c_2$ as $g(x,y,z)=c_1$ and $h(x,y,z)=c_2$, respectively.

First of all, we shall realize that the equations $g(x,y,z)=c_1$ and $h(x,y,z)=c_2$ imply that there are two dependent variables and one independent variable, for example: $y=y(x$) and $z=z(x)$, with $x$ the independent variable.

Taking differentials of $g(x,y,z)=c_1$ and $h(x,y,z)=c_2$, we have $dg=g_x dx+g_y dy+g_z dz=0$ and $dh=h_x dx+h_y dy+h_z dz=0$. This can be arranged as $-g_x dx=g_y dy+g_z dz=0$ and $-h_x dx=h_y dy+h_z dz=0$ which in matrix is

$$ \binom{-g_x dx}{-h_x dx} = \begin{bmatrix} g_y & g_z \\ h_y & h_z \end{bmatrix} \binom{dy}{dz} $$

Here we want to solve solve $dy$ as $dy=G(x,y,z)\ dx$ and $dz$ as $dz=H(x,y,z)\ dx$, where G and H are two yet unknown functions. We could solve it by row operations o Cramer's rule. Using Cramer's rule we get:

$$ dy = \frac{\begin{vmatrix} - g_x\ dx & g_z \\ - h_x\ dx & h_z \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } = - \frac{\begin{vmatrix} g_x & g_z \\ h_x & h_z \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } dx$$

$$ dz = \frac{\begin{vmatrix} g_y & - g_x\ dx \\ h_y & - h_x\ dx \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } = - \frac{\begin{vmatrix} g_y & g_x \\ h_y & h_x \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } dx $$

Rearranging and realizing that the determinants are Jacobians, we get

$$ \frac{dy}{dx} = - \frac{\begin{vmatrix} g_x & g_z \\ h_x & h_z \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } = - \frac{ \frac{\partial(g,h)}{\partial(x,z)} }{ \frac{\partial(g,h)}{\partial(y,z)} } $$

$$ \frac{dz}{dx} = - \frac{\begin{vmatrix} g_y & g_x \\ h_y & h_x \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } = - \frac{ \frac{\partial(g,h)}{\partial(y,x)} }{ \frac{\partial(g,h)}{\partial(y,z)} } $$

Now I will use this to solve the constrained optimization exercise put above, which is:

Optimize $F(x,y,z)=20+2x+2y+z^2$ constrained by $g(x,y,z)=x^2+y^2+z^2=11$ and $h(x,y,z)=x+y+z=3$.

This constraints imply that two variable are dependent and one is independent. For example, $y=y(x)$ and $z=z(x)$. So, using the chain rule $$ \frac{\partial F}{\partial x} = \frac{\partial}{\partial x} \left ( 2x \right ) + \frac{\partial}{\partial y} \left ( 2y \right ) \frac{dy}{dx} + \frac{\partial}{\partial z} \left ( z^2 \right ) \frac{dz}{dx} =2+2\frac{dy}{dx}+2z\frac{dz}{dx}=0$$

To compute $\frac{dy}{dx}$ and $\frac{dz}{dx}$ I use the above result,

$$g_x=2x\ \ \ \ \ \ g_y=2y\ \ \ \ \ \ g_z=2z$$ $$h_x=1\ \ \ \ \ \ h_y=1\ \ \ \ \ \ h_z=1$$

$$ \frac{dy}{dx} = - \frac{\begin{vmatrix} g_x & g_z \\ h_x & h_z \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } = - \frac{\begin{vmatrix} 2x & 2z \\ 1 & 1 \end{vmatrix}}{\begin{vmatrix} 2y & 2z \\ 1 & 1 \end{vmatrix} } = -\frac{2x-2z}{2y-2z}= -\frac{x-z}{y-z} $$

$$ \frac{dz}{dx} = - \frac{\begin{vmatrix} g_y & g_x \\ h_y & h_x \end{vmatrix}}{\begin{vmatrix} g_y & g_z \\ h_y & h_z \end{vmatrix} } = \frac{\begin{vmatrix} 2y & 2x \\ 1 & 1 \end{vmatrix}}{\begin{vmatrix} 2y & 2z \\ 1 & 1 \end{vmatrix} } = - \frac{2y-2x}{2y-2z}=- \frac{y-x}{y-z} $$

So, we get

$$ \frac{\partial F}{\partial x} = 2-2 \frac{x-z}{y-z}-2z \frac{y-x}{y-z}=2\frac{(y-z)-(x-z)-z(y-x)}{y-z}=2\frac{(y-x)-z(y-x)}{y-z}=2\frac{(y-x)(1-z)}{y-z}=0 \ \ \ \ (y\neq z) $$

Solving the system of equations

$$ 2\frac{(y-x)(1-z)}{y-z}=0 \tag{E1} $$ $$ x^2+y^2+z^2=11 \tag{E2} $$ $$ x+y+z=3 \tag{E3} $$

we obtain that the extrema are in the points

$$ \left (x,y,z\right )= \left \{ \left (1+\frac{2}{\sqrt{3}},1+\frac{2}{\sqrt{3}},1-\frac{4}{\sqrt{3}} \right ),\left (1-\frac{2}{\sqrt{3}},1-\frac{2}{\sqrt{3}},1+\frac{4}{\sqrt{3}} \right ),\left (3,-1,1\right ),\left (-1,3,1\right ) \right \} $$

which is correct acording to Wolfram Alpha (the first and second points are maximum and the third and fourth are minimum).

By the way, the system of equations that arise from Lagrange multipliers is

$$2=2\lambda x +\mu $$ $$2=2\lambda y+\mu $$ $$2z=2\lambda z+\mu $$ $$ x^2+y^2+z^2=11 $$ $$ x+y+z=3 $$

In this particular example I think this method (implicit derivative) is easier than Lagrange multipliers. From equation $E1$ it's readily seen that either $x=y$ or either $z=1$, so the system is almost solved.

  • $\begingroup$ With more constraints, the implicit function theorem still goes through, but you have one fewer parameter specifying the surface. For instance, with three variables and two constraints, you have an implicit function $u(x):=\begin{bmatrix} y(x) \\ z(x) \end{bmatrix}$. (Rather, you have possibly several such functions on different neighborhoods, but you already knew about this issue.) $\endgroup$ – Ian Mar 6 '15 at 18:46
  • 1
    $\begingroup$ For instance, one way you could do this in your case would be to write $z=3-x-y$ and then substitute into the other constraint, obtaining $x^2+y^2+(3-x-y)^2=11$. Now you're back to the previous case with two variables and one constraint. Of course usually you cannot explicitly solve either constraint in order to eliminate a variable, but this is the idea. $\endgroup$ – Ian Mar 6 '15 at 18:50
  • $\begingroup$ Thank you for answer. Yes, we could do that, but isn't another way, using the partial derivative equations? $\endgroup$ – user246185 Mar 6 '15 at 18:53
  • $\begingroup$ Using directly equations $(3)$ and $(4)$. $\endgroup$ – user246185 Mar 6 '15 at 18:59
  • $\begingroup$ I'd say the Lagrange multiplier system is still easier. For instance, subtracting the first two equations give either $x=y$ or $\lambda=0$. Substituting $\lambda=0$ into the first equation gives $\mu=2$, then substituting both into the third gives $z=1$. Now you're in pretty much the same situation: either $x=y$ or $z=1,\lambda=0,\mu=2$. $\endgroup$ – Ian Mar 9 '15 at 21:47

If you want to do this with the implicit function theorem, you have to formulate it in the multivariable setting.

You have a function $G : \mathbb{R}^3 \to \mathbb{R}^2$ which defines your constraint. You think of $\mathbb{R}^3$ as $\mathbb{R} \times \mathbb{R}^2$, where the two are the "$y$" variables and the one is the "$x$" variable.

Then $DG(x,y) : \mathbb{R}^3 \to \mathbb{R}^2$ is the Jacobian of $G$ at $(x,y) \in \mathbb{R}^3$. Now $G(x,y)=c$ defines an implicit function $g : \mathbb{R} \to \mathbb{R}^2$, which we think of as $y=g(x)$. $DG$ can be broken up into two blocks, $DG_y \in \mathbb{R}^{2 \times 2}$ from the $y$ derivatives and $DG_x \in \mathbb{R}^{2 \times 1}$ from the $x$ derivatives.

Now the implicit function theorem tells you that

$$\frac{dg}{dx} = -\left ( \left (DG_y \right )(x,g(x)) \right )^{-1} DG_x(x,g(x)).$$

Now you use this with the chain rule to find $\frac{dF}{dx}$, that is, the derivative of $F$ with respect to $x$ while taking into account how $y$ and $z$ respond to changes in $x$. (Here I am slightly lapsing into physicist notation, I hope you will excuse this.) This is given by (choosing $g_1=y,g_2=z$)

$$\frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dg_1}{dx} + \frac{\partial F}{\partial z} \frac{dg_2}{dx}.$$

Now you set that equal to zero.

(Another thing to be excused is that I have written the implicitly defined functions as defined on the whole space when they are really only locally defined. Please excuse this abuse as well; the point is really to label how many free parameters each function has.)

  • $\begingroup$ Although I don't fully understand your answer, it motivated me to search more about the implicit function theorem, and there I found the answer to my question. Now I've edited my question and added the answer to it. Tell me if what I've found is what you were telling in your post. $\endgroup$ – user246185 Mar 9 '15 at 20:43
  • $\begingroup$ @user246185 What you wrote is essentially the same as what I wrote, just written in terms of differentials. $\endgroup$ – Ian Mar 9 '15 at 21:48
  • $\begingroup$ @user246185 Note that you can also write my equation as $\frac{dF(x,g(x))}{dx} = \frac{\partial F}{\partial x} + \nabla_u F \cdot \frac{dg}{dx}$ (where $u$ is the vector $\begin{bmatrix} y \\ z \end{bmatrix}$). Understood this way, all I have really done is write the chain rule and then told you how to calculate $\frac{dg}{dx}$. $\endgroup$ – Ian Mar 9 '15 at 21:53

Let $g:\mathbb{R}^3 \to \mathbb{R}^2$ be given by $g((x,y,z)) = (g_1(x,y,z)-11, g_2(x,y,z)-3)$. The feasible set is described by $g^{-1}(\{0\})$, and you are trying to find $z$ as a function of $x,y$ locally.

Let me abuse notation and write $g(p,z)$ with $p=(x,y)$ to simplify my life.

Then you need to show that ${\partial g(p,z)) \over \partial p}$ is invertible at a solution $(\hat{p}, \hat{z})$, and the implicit function theorem gives some $\pi$ defined locally around $\hat{z}$ such that $g(\pi(z),z) =0$ for $z$ near $\hat{z}$, and we have ${\partial \pi(\hat{z})) \over \partial z} = - {\partial g(\hat{p},\hat{z})) \over \partial p}^{-1} {\partial g(\hat{p},\hat{z})) \over \partial z}$.

If we let $\phi(z) = f(\pi(z),z)$ (abusing notation again), then at a solution we will have ${\partial \phi(\hat{z})) \over \partial z} = 0$, using the chain rule this reduces to ${\partial f(\hat{z},\hat{p})) \over \partial z} = {\partial f(\hat{z},\hat{p})) \over \partial p} {\partial g(\hat{p},\hat{z})) \over \partial p}^{-1} {\partial g(\hat{p},\hat{z})) \over \partial z}$.

(As an aside, when used as a method for numerical optimization, this is known as 'reduced gradients'.)


Extremising functions which satisfy one or more constraints is the topic of Lagrange multipliers (see book 'Calculus' by Stewart, chapter 15.8, for instance).

To extremise a function $f(x,y,z)$ subject to two constraints $g(x,y,z) = k$ and $h(x,y,z) = c$ we need to solve the following equation: $\nabla f(x_{0},y_{0},z_{0}) = \lambda \nabla g(x_{0},y_{0},z_{0}) + \mu \nabla h(x_{0},y_{0},z_{0})$, where $\lambda$ and $\mu$ are called Lagrange multipliers and the point $(x_{0},y_{0},z_{0})$ is a solution point.

Essentially, this means solving the following five equations: $f_{x} = \lambda g_{x} + \mu h_{x}, f_{y} = \lambda g_{y} + \mu h_{y}, f_{z} = \lambda g_{z} + \mu h_{z}, g(x,y,z) = k, h(x,y,z) = c$.

We have five equations, five unknowns. Of course, these may not be easily solved, but the usual procedure is to use the first three equations to find the Lagrange multipliers $\lambda$ and $\mu$ (possibly in terms of $x,y,z$), then use the last two equations to find the values of $x,y,z$. Finally, evaluate $f$ at the values $x,y,z$.

There is a nice geometrical explanation of where the above equation comes from, again I refer to Stewart, section 15.8.

EDIT: sorry, I did not realise you were looking for a method using the Implicit Function Theorem.

  • $\begingroup$ The OP already mentioned that (s)he knows how to do it with Lagrange multipliers and would like to know how one might do it with implicit functions. $\endgroup$ – Ian Mar 6 '15 at 18:54

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