Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A region is given by , , and . Its corners are , , and . The objective function is , where is a positive number. Find the value of for which is the same at and at . For that , find the maximum of over the region and the number of corners where it occurs.
For a bounded feasible region, work out at every corner point. The largest value is the maximum and the smallest value is the minimum. State the best point as well as the best value.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Set the two values equal. , so and .
Step 2. The values for . At , . At , . At , . At , .
Step 3. Compare. The maximum is , and it occurs at two corners, and . Every point of the edge between them also gives .
Answer: , the maximum of is , and it occurs at corners.
What if? For a little below the best corner is , and for a little above it is . The tie is where the best corner changes.
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