We are encountering unusual behaviour with one of the Parsons problems in APEX, and I can't figure out what is wrong.
The question is here:
https://runestone.academy/ns/books/published/APEXPROTEUS/sec_optimization.html#APEX-PROTEUS-optimization-2-v1
I'll post the source below.
For some students, the blocks do not show up on the assignment page. When they view the question in the book, the blocks show up, and they're able to complete the problem.
But when they return to the assignment page, the check mark that should indicate a successful attempt is not there. It seems that the Runestone grader is giving credit, but students are understandably concerned.
When I tried it, the blocks did show up on the assignment page. I solved it, and the green check mark showed up.
But then I viewed the problem in the book. Two of the seven blocks were missing in the solution, and the problem had the yellow warning sign instead of the green check.
I clicked the reset button, the blocks came back, and I was able to solve again while viewing in the book.
But when I returned to the assignment page, the blocks were gone completely, and there was no indication that the problem had been attempted, successfully or unsuccessfully. This time, clicking reset did not bring back the blocks.
Here is the PreTeXt source for the problem:
<exercise label="APEX-PROTEUS-optimization-2-v1" adaptive="yes" component="proteus">
<title>Parsons Problem, Optimization</title>
<statement>
<p>
Follow <xref ref="idea_optimization"/> to solve the following optimization problem:
What are the dimensions of a cylinder of a given
volume <m>V</m> so that its total surface area <m>S</m> is minimal?
</p>
<image width="33%">
<shortdescription>A diagram of a right circular cylinder, labeling the radius and height.</shortdescription>
<description>
<p>
A generic right circular cylinder is drawn without reference to a coordinate system.
The cylinder is oriented vertically, with a circular base at the bottom.
</p>
<p>
One side of the cylinder is labeled <m>h</m>, for the height,
and a line segment is drawn from the center of the circular top to the edge,
and labeled <m>r</m>, for radius.
</p>
</description>
<latex-image label="proteus-img_right_circ_cyl">
\begin{tikzpicture}[x=13pt,y=14pt,thick,scale=1.13]
\begin{scope}[xscale=2]
\draw (-1,0) arc (-180:0:1);
\draw [dashed] (1,0) arc (0:180:1);
\draw (0,2.5) circle (1);
\draw (-1,0) -- (-1,2.5) (1,0)-- (1,2.5) node [right,pos=.5] {\(h\)};
\draw (0,2.5) -- (1,2.5) node [above,pos=.5] {\(r\)};
\end{scope}
\draw [fill=black] (0,2.5) circle (1pt);
\end{tikzpicture}
</latex-image>
</image>
</statement>
<blocks>
<block order="1">
<p>
The quantity to minimize is the total surface area of the cylinder.
</p>
</block>
<block order="2">
<p>
The <em>fundamental equation</em> is <m>S= 2\pi r^2 + 2\pi r h</m>
</p>
</block>
<block order="3">
<!-- <choice> -->
<p>
The <em>constraint equation</em> is <m>V = \pi r^2 h</m>.
</p>
</block>
<block order="5">
<p>
Then, the <em>reduced fundamental equation</em> is <m> S(r) = 2\pi r^2 + 2\pi r \dfrac{V}{\pi r^2} = 2\left(\pi r^2 + \dfrac{V}{r}\right)</m>.
</p>
</block>
<block order="4">
<p>
The domain of the reduced fundamental equation is <m> 0 \lt r \lt \infty </m>
</p>
</block>
<block order ="5">
<p>
The <em>extreme value</em> of the reduced equation is <m>r = \left(\dfrac{V}{2\pi}\right)^{\frac{1}{3}}</m>
</p>
</block>
<block order="6">
<p>The cylinder of volume <m>V</m> with minimal total surface are area has radius
<m>r =\left(\frac{V}{2\pi}\right)^{\frac{1}{3}}</m> and height <m>h=2r</m>. That is, the height of the cylinder is equal to its diameter.
</p>
</block>
</blocks>
</exercise>