Determine graphically the solution set for each system of inequalities and indicate whether the

Determine graphically the solution set for each system of...

Key Concepts

Bounded vs Unbounded Regions

Once the feasible region is identified, it is important to determine whether it is bounded (completely enclosed within a finite area) or unbounded (extending indefinitely in one or more directions). This distinction can affect the existence and nature of optimal solutions in optimization problems.

Graphical Representation of Inequalities

This concept involves plotting each inequality on a coordinate plane by first converting the inequality into an equation to draw its boundary line, and then determining which side of the line represents the solution set of the inequality. It is a crucial technique in understanding the spatial relationships between the different constraints of a system of inequalities.

Half-Plane Analysis

Each linear inequality divides the plane into two half-planes. Recognizing and shading the correct half-plane for each inequality is essential. By overlaying these half-planes, one can visually determine which portions of the plane satisfy all the inequalities simultaneously, leading to the feasible region.

Feasible Region

The feasible region is the intersection of all the half-planes defined by the inequalities. It represents all the points that satisfy every inequality in the system. This concept is fundamental in linear programming and optimization, as it defines the set of possible solutions to a problem.

Intersection of Constraints

Understanding that the solution to a system of inequalities is the common intersection of all individual constraints is key. This concept emphasizes that every point in the feasible region must satisfy all conditions imposed by the inequalities, including any restrictions like non-negativity constraints (e.g., x ? 0, y ? 0).

Transcript

00:02 Okay, so in this problem, we are asked to find the graph of this system of inequalities that is made of y is greater than or equal to x square and x plus y is greater than or equal to six.

00:14 We're also asked to find whether or not the solution set is bounded and to find all vertices.

00:20 To start, we're going to set these two equations to equal to each other to find the vertex or the interception of these two, which can or not be part of our solution set.

00:32 So we're going to start out by stating that y is equal to 6 minus x and we're going to set this equal to x square.

00:42 Oh, sorry.

00:44 So this is going to be equals to x square.

00:48 Now, we're going to take all of this on one side.

00:52 So x square is equal to, sorry, x square plus x minus x.

00:58 It's going to be equal to zero.

01:02 Now, we can determine that x, this is a quadratic equation.

01:11 So we can find these solutions for this two, for this equation over here by factoring out two terms.

01:18 We can see, we can clearly see that x has a coefficient of one.

01:23 So we have two x, two xes that are going to be multiplied by each other.

01:27 And so we're looking for numbers that multiplied by each other.

01:35 Give minus 6, and that added give a plus 1.

01:47 So these two numbers must have a different sign since we're looking for a negative value and a positive value when adding.

01:57 So a solution for this will be to state that negative 2 times 3 is negative 6, and we can try out and see whether or not this fits this addition.

02:08 3 plus negative 2 is indeed 1.

02:12 So we have the solution and we have factored out this two, this quadratic equation.

02:19 So we have x, x minus two times x plus three is equal to zero.

02:33 Now we can take these two factors and say that x can be either x minus two can be either zero or x plus three can be zero.

02:47 And either of these two statements will make this whole equation over here equal to zero.

02:55 So we can have that x can be 2 or x can be negative 3.

03:03 And we can substitute this back into the equation to one of these first equations.

03:08 I'm going to go ahead and just take these two and substitute them into the equation y is greater than or equal to x square.

03:16 So we're going to have y is equal to two a square.

03:22 Or y is equal to negative 3 square.

03:29 For this value, we'll have that y is equal to four.

03:37 And for this one, we'll have that y is equal to nine.

03:41 So we have two points in which these two lines intersect.

03:45 The point 2 .4 and the point negative 3 .9.

03:58 And these are one of our vertices for this solution.

04:02 Of inequalities.

04:04 Finally, i'm just going to take a look at this equation right here, and we're going to have two points that are going to be useful to draw this line.

04:14 So we have the line x plus y is equal to six.

04:19 Now, subtracting x on both sides of the equation will have that 6 minus x is equal to y.

04:28 So this should be 6 minus x is equal to y.

04:33 And so we have a y intercept at 6 -0 .6, sorry.

04:44 When x is equal to 0, we have a y intercept of 6 .0.

04:48 And we have an x intercept when y is equal to 0.

04:53 We just solved for x.

04:55 We'll have that 6 .0 is the next intercept of this equation.

05:02 And we know how to grab the equation y is equal to x square...