In Exercises 5–14, an objective function and a system of linear i...

Question

In Exercises 5–14, an objective function and a system of linear inequalities representing constraints are given. a. Graph the system of inequalities representing the constraints. b. Find the value of the objective function at each corner of the graphed region. c. Use the values in part (b) to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.

Objective function and constraints for graphing systems of inequalities in college algebra.

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of inequalities and were asked to do several things with it. First, we need to graph it, then we need to evaluate the objective function at each corner of the graft region. The objective function here is this the equals X plus eight Y. And then finally, we are going to identify the values of X and Y for which the maximum value is found. So quite a few steps here. Well, let's start off with the graphing step. We're going to draw a rough sketch of a coordinate plane. We're going to have a horizontal X axis of vertical Y axis and they come together the origin in the middle. For the first two inequalities that were given the first one is X is greater than or equal to zero. Where is that true? Where is X greater than or equal to zero? Well, that's anywhere to the right of and including the Y axis. That's all the places where X is positive or greater than or equal to zero. We're also given the inequality, why is greater than or equal to zero? Where is that true. Well, that's going to be anywhere above and including the X access. That's where, why is going to be positive. So where is it true that both X and Y are greater than, or equal to zero? That's going to be the first quadrant here. And so what I'm going to do is draw another rough sketch of a coordinate plane. But this time really zoomed in on the first quadrant because this is the region where at least the first two sets of the first two inequalities are true. And we need to figure out where within that first quadrant are the second to inequality is true. So let's start working on the inequality of six X plus five Y is greater than or equal to 30. Well, we recall from previous lessons that to graph an inequality, we are first going to write it as an equality. So we've got six X plus five Y equals 30. We recognize that this is an equation of a line. So we can graph the intercepts starting off with the Y intercept where X is equal to zero, plugging zero in four, our X value. So six times not X but zero plus five Y equals 36 times zero is zero. So we have five Y equals 30 while dividing both sides by five that gives us a Y value of six. So looking at our graph starting from the origin 00, we are going neither left nor right for our X value of zero. And we're going up six units for our Y value of six. So we can label this 06. Now let's plot our X intercept for this line plug zero in for the Y value. So we have six X plus five, not times Y, but now times zero equals 35 times +00. So we have six X equals 30 dividing both sides by six, we get X equals five. Now we can plot the 50.50 from the origin. We're moving to the right five units for our X value of five and neither up nor down for our Y value of zero. Now, we can connect these 2.6250. And now because this is an inequality, we want to figure out if we're talking about the region that's down And to the left of this line or up and to the right of this line. And taking note that the line is included in our inequality because it is greater than, or equal to. And all four of these inequalities include the or equal to parts. So they're all going to be solid lines. But let's choose a test point. I'm going to choose the test zero for this inequality. And that will tell us if we're talking about down into the left of this line where it's true. So let's plug in 00 for X and Y values. So six times. Not expert, zero plus five times. Not why, but zero is greater than, or equal to 30. Well, six times 00 plus five times 00. That's all zero. On the left hand side. Is it true that zero is greater than, or equal to 30? No, that is not true. So we can reject the test zero and reject that whole region down into the left. We're talking about the region up into the right of our line. So this is our new region that we are taking a look at. Now, we've got one more inequality to graph. We have four, X plus five Y is less than, or equal to 40. And again, we put that as an equality as an equation. Four X plus five Y equals 40. Let's graph the intercepts for this line. So the Y intercept plugging zero in for X four times not X but zero plus five, Y equals 44 times zero is zero. So we have five, Y equals 40 dividing both sides by five, we get Y equals eight. So plotting the .08 from the origin, we are going neither left nor right for our x value of zero and growing up eight units for our y value of eight and plotting this .08. Now let's get our X intercept. So plaguing zero in for Y, we have four times X plus five times not Y but zero equals 40 and five times zero is zero. So we have four, X equals 40 dividing both sides by four, we get X equals 10. So plotting the 100.10 0 from the origin, going to the right 10 units for our X value of 10 and neither up nor down for our Y value of zero. So we can label this 10 0. Now, we can draw a line connecting those two points from 08 to 10 0. And now we need to know if we're talking about the region that is down to the left of this line or up and to the right of this line. And we can choose a test point. Our test point does not have to be within that region that would fit all of the inequalities. It just has to be below or above the particular inequality that we're talking about. So I'm going to choose 00 again because that's clearly down into the left. So plugging in 00, we've got four times, not expert zero plus five times, not Y but zero is less than, or equal to 40 four times 005 times 00. So on the left, that's just zero again. Is it true that zero is less than, or equal to 40? Yes, it is zero is less than or equal to 40. So we are talking about that region that is down and to the left of our line. So narrowly in between these two lines, this is the region we're talking about. And because we are including the X and Y axes, we can connect those lines from 08 to 06 on the Y axis and from 50 to 10 0 on the X axis. And here is our beautiful graft region. The next step now that we've graphed, it is to evaluate the objective function at each corner of the graft region. So in order to do that, we're taking that objective function, Z equals X plus eight Y And we are going to plug in the X&Y value for each of the corners in turn. So let's start with 06. So plugging in zero for X that will be zero plus and plugging in six for Y, that will be eight times not Y but six, well, zero plus is nothing and eight times six is 48. So the Z value at 06 is 48. Now let's do the 480. plugging in zero for X that's zero plus plugging in eight for Y, that's eight times eight, zero plus is nothing. eight times 8 is 64. Now let's look at the corner for 10 0, plugging in 10 for X, that's 10 plus and then plugging in zero for Y that's eight times not why, but zero, eight times zero is zero. So this is just 10 And the last corner 50 plugging in five for the X value plus eight times and zero again for the Y value eight times zero is still zero. So this is five. Alright, now we just need to determine the maximum value. So looking for the Z values 48 64 10 and five, what's the maximum was the highest one? That's 64? So when X equals zero and Y equals eight, we're at our maximum value. Wonderful. We have found all of our answers. Now, we just need to find the appropriate answer choice. So first, we want to look for corners at 0608, 10 0 and 50. And both A and B have those corners. Great. Those graphs look good. Now, let's look at these Z values for the corner of 06. We want a Z value of 48. Well, a has 20, that's not going to work, but B does have a z value at 48 for the .06. Great. Now, how about 408? We need a value of 64. We've got that now for 10 0, a value of 10. Yes. Answer choice B is doing well and for 50 and Z value of five. Yes. Thank you very much. Answer choice B A maximum value at 64 where X equals zero and Y equals eight. That's what we got. Answer Choice B works well done. We'll catch you on the next one.

Key Concepts

Linear Inequalities

Graphing Systems of Inequalities

Objective Function

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