Given the question:
Mountain Mark and Big Lake Bob spend their time carving fishing lures and duck decoys. The table below shows their production possibility schedule, which describes various combinations of fishing lures and duck decoys they each can carve efficiently in a week.
[tex]\begin{center}
\begin{tabular}
{|c {2.3cm}|c {2.3cm}|c {2.3cm}|c {2.3cm}|}
\multicolumn {4} {|c|} {Production Possibility Schedules}\\[1ex]
\multicolumn {2} {|c|} {Mountain Mark}&\multicolumn {2} {|c|} {Big Lake Bob}\\[1ex]
Fishing Lures&Duck Decoys&Fishing Lures&Duck Decoys\\[1ex]
40&0&100&0\\
32&10&80&6\\
24&20&60&12\\
16&30&40&18\\
8&40&20&24\\
0&50&0&30
\end{tabular}
\end{center}[/tex]
If
mountain mack splits his week evenly between carving fishing lures and
duck decoys, what is the maximum number of fishing lures and duck decoys
he could carve?
From the table it can be seen that in a full week, if mountain mack is carving fishing lures only, he can carve 40 fishing lures. This means that in half of a week, he can carve 20 fishing lures.
Similarly, from the table it can be seen that in a full week, if mountain
mack is carving duck decoys only, he can carve 50 duck decoys. This
means that in half of a week, he can carve 25 duck decoys.
Therefore, if
mountain mack splits his week evenly between carving fishing lures and
duck decoys, the maximum number of fishing lures and duck decoys
he could carve are 20 fishing lures and 25 duck decoys.
