In Problems 9–18, solve each linear programming problem. Minimize z=5 x+4 y subject to x

In Problems 9–18, solve each linear programming problem. ...

Key Concepts

Objective Function

The objective function is the equation that needs to be optimized (either minimized or maximized) in a linear programming problem. It is a linear function of the decision variables and represents the goal of the problem, such as minimizing cost or maximizing revenue.

Corner Point Principle

The corner point principle, also known as the vertex theorem, states that for a linear programming problem, the optimal solution (if one exists) is located at one of the vertices or corner points of the feasible region. This principle is useful because it allows one to focus on a finite number of candidate solutions when seeking the optimum.

Feasible Region

The feasible region is the set of all points that satisfy all the constraints of a linear programming problem. It represents all potential solutions where the decision variables meet the required conditions, and the optimal solution is usually found at the boundary of this region.

Linear Programming

Linear programming is an optimization method used to find the best outcome, such as maximum profit or minimum cost, in a mathematical model whose requirements are represented by linear relationships. It involves an objective function that is optimized (maximized or minimized) subject to a set of linear inequality or equality constraints.

Constraints

Constraints are linear inequalities or equations that define the restrictions or limits within which the decision variables must operate. They represent the conditions that must be satisfied for a solution to be considered feasible in a linear programming problem.

Transcript

00:01 So we're given our objective function as well as our constraints here, and we want to start solving this linear programming problem by graphing the constraints.

00:11 So given that x and y are both greater than or equal to 0, we know that our graph will lie in the first quadrant.

00:19 And now we're going to start grapping these lines down here, and we can start by rearranging this function to be y is greater than or equal to negative x plus 2.

00:27 Now i can graph this with the y intercept of 2, a slope of negative 1, meaning that the x intercept will also be 2.

00:36 Now this following line, i can rearrange as 3y is less than or equal to negative 2x plus 12.

00:44 I divide both sides by 3 and i end up with y is less than or equal to negative 2 thirds x plus 4.

00:53 So our y intercept is at 4.

00:55 We have a slope of negative 2 .2.

00:57 3rds, meaning that our x intercept would be at 6.

01:03 Now for this final equation, i can rearrange it to be y is less than or equal to negative 3x plus 12.

01:11 So our y intercept would be at 12, and because our slope is negative 3, our x intercept is at 4, and the line looks like this.

01:23 And now we want to shade the area where our feasible points lie.

01:29 Which would be here.

01:32 Our next step would be to find the coordinates of all the corner points.

01:37 So starting here we can tell that this corner point would be 0 2.

01:43 This corner point is 0 .4.

01:47 This corner is 2.

01:49 This corner is 2 .0.

01:51 This here is 4 comma 0 and our final corner point we're going to have to do some solving.

01:58 So we want to take this line and this line and take their equations and solve for each other.

02:07 So we're going to set them equal to each other, which would be this line as well as this equation...

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