Video Transcript
Given the graph below and that π₯ is greater than or equal to zero, π¦ is greater than or equal to zero, π₯ plus π¦ is less than or equal to seven, and π¦ is greater than or equal to five, determine at which point the function π equals three π₯ minus π¦ has its maximum using linear programming.
Okay, in this example, weβre given these four constraints, and those constraints are depicted in our figure. Itβs the orange-shaded region that shows us the allowable values for π₯ and π¦ given our restrictions. Along with this, weβre given this function π equals three π₯ minus π¦. We can call this our objective function. This is the function we want to maximize within our given constraints. The idea, then, is that we take our objective function and we plug in π₯, π¦ coordinate pairs where those pairs, the π₯, π¦ points, are taken from the orange-shaded region in our graph.
The point from this area that gives us the largest value of π compared to all the other values will be the point that maximizes this objective function. Now, in this orange-shaded region, there are actually infinitely many points. But thankfully, due to a proof that we wonβt give here, itβs only the points at the vertices of this region β at the points labeled π΄, π΅, and πΆ β which could lead to a maximum value in our objective function. This means itβs really only these three points that we need to test.
Note that thereβs a fourth point labeled on our graph called point π·. Because this point is outside of the allowed or feasible region though, the one shaded in orange, we wonβt consider it. Our next step, then, is to figure out the coordinates of these three points π΄, π΅, and πΆ. For point π΄, we can see that this lies along the line π₯ equals zero and the line π¦ equals five. Point π΅ has the same π¦-value as point π΄ with an π₯-value of two. Point πΆ has an π₯-value of zero and a π¦-value of seven. So then, these are the three points weβll test in our objective function.
Substituting in the coordinates of point π΄, we have three times zero minus five. Thatβs negative five. Then for point π΅, we have three times two minus five. Thatβs positive one. And lastly, point πΆ, three times zero minus seven equals negative seven. Of these three points, the maximum value is positive one. And that corresponds to the coordinates of point π΅.
We can say then that itβs at point π΅ at which our objective function reaches its maximum value.