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diff --git a/source/know/concept/lagrange-multiplier/index.md b/source/know/concept/lagrange-multiplier/index.md
index 9761e75..8ee1054 100644
--- a/source/know/concept/lagrange-multiplier/index.md
+++ b/source/know/concept/lagrange-multiplier/index.md
@@ -10,27 +10,27 @@ layout: "concept"
 
 The method of **Lagrange multipliers** or **undetermined multipliers**
 is a technique for optimizing (i.e. finding the extrema of)
-a function $f(x, y, z)$,
-subject to a given constraint $\phi(x, y, z) = C$,
-where $C$ is a constant.
+a function $$f(x, y, z)$$,
+subject to a given constraint $$\phi(x, y, z) = C$$,
+where $$C$$ is a constant.
 
-If we ignore the constraint $\phi$,
-optimizing $f$ simply comes down to finding stationary points:
+If we ignore the constraint $$\phi$$,
+optimizing $$f$$ simply comes down to finding stationary points:
 
 $$\begin{aligned}
     0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z}
 \end{aligned}$$
 
 This problem is easy:
-$\dd{x}$, $\dd{y}$, and $\dd{z}$ are independent and arbitrary,
+$$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$ are independent and arbitrary,
 so all we need to do is find the roots of
-the partial derivatives $f_x$, $f_y$ and $f_z$,
-which we respectively call $x_0$, $y_0$ and $z_0$,
-and then the extremum is simply $(x_0, y_0, z_0)$.
+the partial derivatives $$f_x$$, $$f_y$$ and $$f_z$$,
+which we respectively call $$x_0$$, $$y_0$$ and $$z_0$$,
+and then the extremum is simply $$(x_0, y_0, z_0)$$.
 
-But the constraint $\phi$, over which we have no control,
-adds a relation between $\dd{x}$, $\dd{y}$, and $\dd{z}$,
-so if two are known, the third is given by $\phi = C$.
+But the constraint $$\phi$$, over which we have no control,
+adds a relation between $$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$,
+so if two are known, the third is given by $$\phi = C$$.
 The problem is then a system of equations:
 
 $$\begin{aligned}
@@ -42,15 +42,15 @@ $$\begin{aligned}
 Solving this directly would be a delicate balancing act
 of all the partial derivatives.
 
-To help us solve this, we introduce a "dummy" parameter $\lambda$,
+To help us solve this, we introduce a "dummy" parameter $$\lambda$$,
 the so-called **Lagrange multiplier**,
-and contruct a new function $L$ given by:
+and contruct a new function $$L$$ given by:
 
 $$\begin{aligned}
     L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z)
 \end{aligned}$$
 
-At the extremum, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$,
+At the extremum, $$\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$$,
 so now the problem is a "single" equation again:
 
 $$\begin{aligned}
@@ -58,13 +58,13 @@ $$\begin{aligned}
     = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z}
 \end{aligned}$$
 
-Assuming $\phi_z \neq 0$, we now choose $\lambda$ such that $f_z + \lambda \phi_z = 0$.
+Assuming $$\phi_z \neq 0$$, we now choose $$\lambda$$ such that $$f_z + \lambda \phi_z = 0$$.
 This choice represents satisfying the constraint,
-so now the remaining $\dd{x}$ and $\dd{y}$ are independent again,
-and we simply have to find the roots of $f_x + \lambda \phi_x$ and $f_y + \lambda \phi_y$.
+so now the remaining $$\dd{x}$$ and $$\dd{y}$$ are independent again,
+and we simply have to find the roots of $$f_x + \lambda \phi_x$$ and $$f_y + \lambda \phi_y$$.
 
-In effect, after introducing $\lambda$,
-we have four unknowns $(x, y, z, \lambda)$,
+In effect, after introducing $$\lambda$$,
+we have four unknowns $$(x, y, z, \lambda)$$,
 but also four equations:
 
 $$\begin{aligned}
@@ -73,19 +73,19 @@ $$\begin{aligned}
     \phi = C
 \end{aligned}$$
 
-We are only really interested in the first three unknowns $(x, y, z)$,
-so $\lambda$ is sometimes called the **undetermined multiplier**,
+We are only really interested in the first three unknowns $$(x, y, z)$$,
+so $$\lambda$$ is sometimes called the **undetermined multiplier**,
 since it is just an algebraic helper whose value is irrelevant.
 
 This method generalizes nicely to multiple constraints or more variables:
-suppose that we want to find the extrema of $f(x_1, ..., x_N)$
-subject to $M < N$ conditions:
+suppose that we want to find the extrema of $$f(x_1, ..., x_N)$$
+subject to $$M < N$$ conditions:
 
 $$\begin{aligned}
     \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M
 \end{aligned}$$
 
-This once again turns into a delicate system of $M+1$ equations to solve:
+This once again turns into a delicate system of $$M+1$$ equations to solve:
 
 $$\begin{aligned}
     0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N}
@@ -97,15 +97,15 @@ $$\begin{aligned}
     0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N}
 \end{aligned}$$
 
-Then we introduce $M$ Lagrange multipliers $\lambda_1, ..., \lambda_M$
-and define $L(x_1, ..., x_N)$:
+Then we introduce $$M$$ Lagrange multipliers $$\lambda_1, ..., \lambda_M$$
+and define $$L(x_1, ..., x_N)$$:
 
 $$\begin{aligned}
     L = f + \sum_{m = 1}^M \lambda_m \phi_m
 \end{aligned}$$
 
-As before, we set $\dd{L} = 0$ and choose the multipliers $\lambda_1, ..., \lambda_M$
-to eliminate $M$ of its $N$ terms:
+As before, we set $$\dd{L} = 0$$ and choose the multipliers $$\lambda_1, ..., \lambda_M$$
+to eliminate $$M$$ of its $$N$$ terms:
 
 $$\begin{aligned}
     0 = \dd{L}