Question

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?

Answer 1

To find the maximum number of fishing lures and duck decoys Mountain Mack can carve, we need to divide his available time equally between the two activities. The maximum number of each item he can carve depends on the total hours available, the time it takes to carve a fishing lure, and the time it takes to carve a duck decoy.

Step-by-step explanation:

To find the maximum number of fishing lures and duck decoys that Mountain Mack could carve, we need to split his week evenly between the two activities. Let's say he has a total of 'x' hours available in a week. Since he'll be dividing his time equally, he'll spend 'x/2' hours carving fishing lures and 'x/2' hours carving duck decoys.

To put this in terms of fishing lures and duck decoys, we need to know the time it takes to carve each item. Let's say it takes 'y' hours to carve a fishing lure and 'z' hours to carve a duck decoy. The maximum number of fishing lures Mountain Mack can carve would be 'x/2y' and the maximum number of duck decoys he can carve would be 'x/2z'.

So, to find the maximum number of fishing lures and duck decoys Mountain Mack can carve, we need to know the total number of hours available in a week ('x'), the time it takes to carve a fishing lure ('y'), and the time it takes to carve a duck decoy ('z').

Answer 2

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.


`\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} {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}`

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.