fix: typos

src/opt.typ

@@ -66,7 +66,7 @@ Later on we'll do more examples where Step 2 and Step 3 are more intricate.
   so we're in the easy case and the recipe applies here.
   We check all the points in turn:
 
-  0. $cal(R)$ has no boundary and limit cases when any variable approaches $0$ or $+oo$.
+  0. $cal(R)$ has no boundary and limit cases when any variable approaches $0^+$ or $+oo$.
   1. To find the critical points, calculate the gradient
     $ nabla f (x,y) = vec(1 - 8 / (x^2 y), 1 - 8 / (x y^2)) $
     and then set it equal to $vec(0,0)$.

src/regions.typ

@@ -92,7 +92,7 @@ The three that you should care about for this class are the following:
 ]
 
 - The *boundary* is usually the points you get when you choose any one of the $<=$ and $>=$
-  constraints and turn it into and $=$ constraint.
+  constraints and turn it into an $=$ constraint.
   For example, the boundary of the region cut out by $-1 <= x <= 1$ and $-1 <= y <= 1$
   (which is a square of side length $2$)
   are the four sides of the square, where either $x = pm 1$ or $y = pm 1$.

src/sol-foxtrot.typ

@@ -157,7 +157,7 @@ Now we check the cases:
   So every lattice point is indeed a saddle point.
 
 - At any point of the form $(m + 1/2, n + 1/2)$, we have $A = C = pm pi^2$ and $B = 0$,
-  so $A C - B^2 = pi^4 < 0$.
+  so $A C - B^2 = pi^4 > 0$.
   Hence there are no saddle points here.
 
 == Solution to @exer-opt-geo (geometry optimization)