diff --git a/optimization/exercises/singVarCalcOptimization3.tex b/optimization/exercises/singVarCalcOptimization3.tex
index a7843caaf..d7f26655c 100644
--- a/optimization/exercises/singVarCalcOptimization3.tex
+++ b/optimization/exercises/singVarCalcOptimization3.tex
@@ -34,11 +34,11 @@
   \end{tikzpicture}
   \end{hint}
   \begin{hint}
-  The dimensions that will result in  "least material" for the five sides are dimensions that result in "smallest surface area".
-  So, the quantity that we have to "minimize" is the surface area, $S$.
+  The dimensions that will result in  ``least material'' for the five sides are dimensions that result in ``smallest surface area''.
+  So, the quantity that we have to ``minimize'' is the surface area, $S$.
   \end{hint}
   \begin{hint}
-  We have to "minimize"  the surface area, $S$.
+  We have to ``minimize''  the surface area, $S$.
   
   $S=x^2 +4xy$
   
@@ -64,9 +64,8 @@
   \begin{hint}
   So, we have to find the (global) minimum of $S$ on its domain, $(0,\answer{\infty})$.
   We have to find critical points of $S$.
-  Therefore, we have to compute $S'(x)$.
-  \end{hint}
-  \begin{hint}
+  Therefore, we have to compute $S'(x)$:
+ 
   $S'(x)=2x-\frac{400}{\answer{x^2}}$.
     \end{hint}
      \begin{hint}
@@ -76,9 +75,7 @@
   $2x-\frac{400}{\answer{x^2}}=0$.
   
   It follows that $S$ has the only critical point  $x=\answer{200^{\frac{1}{3}}}$.
-    \end{hint}
-  
-     \begin{hint}
+
      Since
     
      $S'(x)=2x-\frac{400}{x^{2}}=\frac{2(x^{3}-200)}{x^{2}}$, it follows that