Video: Finding the Point That Maximize the Objective Function given the Graph of the Constraints

Using linear programming, find the minimum and maximum values of the function 𝑝 = 4π‘₯ βˆ’ 3𝑦 given that π‘₯ β‰₯ 0, 𝑦 β‰₯ 0, π‘₯ + 𝑦 ≀ 9, and 𝑦 β‰₯ 5.

03:22

Video Transcript

Using linear programming, find the minimum and maximum values of the function 𝑝 equals four π‘₯ minus three 𝑦 given that π‘₯ is greater than or equal to zero, 𝑦 is greater than or equal to zero, π‘₯ plus 𝑦 is less than or equal to nine, and 𝑦 is greater than or equal to five.

We have four linear inequalities in the variables π‘₯ and 𝑦, and these are our constraints. As well as the text of the question, we’re also given a graph where the feasible region has been shaded; that is the set of all the feasible solutions. This shaded region shows where the constraints hold. This is a set of feasible solutions which is the domain to our objective function that we have to maximize or minimize, or in this question both.

In our problem, the feasible region is a triangle. In general, the feasible region will be a polygon for a linear programming problem. And the question asks us to find the extreme values of our objective function 𝑝 equals four π‘₯ minus three 𝑦 in this feasible region; that is, where π‘₯ and 𝑦 are the π‘₯ and 𝑦 coordinates of a point in the feasible region. And you might think this is going to be difficult because we’re going to have to check the value of this function for every single point in this triangle, but actually we only have to check the vertices.

The maximum and minimum values of a linear objective function, like the one we have, are always attained at the vertices of the feasible region. We just have to evaluate this function therefore at the three vertices. So what is the value of the function at the vertex zero, nine? Our objective function 𝑝 is four π‘₯ minus three 𝑦. Our value of π‘₯ is zero, and our value of 𝑦 is nine. So we get four times zero minus three times nine, which is negative 27.

It’s the same process for the vertex zero, five. Our objective function is four π‘₯ minus three 𝑦. And substituting the values of π‘₯ and 𝑦, we get that the value of the objective function at this point is negative 15. And finally, we have the vertex four, five for which the objective function is four times four minus three times five, which is one.

The values of the objective function at the vertices are negative 27, negative 15, and one. Of these three values, the minimum value is negative 27 and the maximum value is one. And these are not just the minimum and maximum values on the vertices of the feasible region; they are the minimum and maximum values on the entire feasible region including the interior and the edges.

And so the minimum and maximum values of the function 𝑝 equals four π‘₯ minus three 𝑦 given that π‘₯ is greater than or equal to zero, 𝑦 is greater than or equal to zero, π‘₯ plus 𝑦 is less than or equal to nine, and 𝑦 is greater than or equal to five are negative 27 and one.

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