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 we need to do several things with it. First, we need to graph it and then next, we need to evaluate the objective function at each corner of the graft region. And then finally, we will identify the values of X and Y for which the maximum value is found. So a lot of steps here, let's start off with the graphing. I'm going to draw a rough sketch of a coordinate plane will have a horizontal X axis of vertical Y axis and they come together at the origin in the middle. Now, for graphing our inequalities, we'll start with the 1st 13 is less than or equal to X, which is less than or equal to seven. And what does this mean? While we're talking about a region between and including three and seven for our possible X values. So from the origin, we can move to the right three units to the .30. And then we can move further to the right to the .70. And we can draw vertical lines through both of those points and they'll be solid vertical lines because these are all less than or equal to. And we're talking about this shaded region here in between those two lines where X equals three and X equals seven. Alright, so we've got the first inequality. Now let's figure out the second one. The second one is why is greater than or equal to four. So we're talking about why values that can be for all the way up to positive infinity. And we show that on the graph starting at the origin, we can move up four units to the .04. And then we can draw a horizontal line through that point and including that point and then going all the way up to positive infinity is the region that we're talking about for why is greater than or equal to four. So now we are talking about the region that has to be included for both of those inequalities. And that's going to be up here in the first quadrant where we have the red and the blue colored in up here. So we can create another graph where we're kind of zoomed in on that first quadrant. So we still have the horizontal X axis, the vertical Y axis, they come together at the origin in the middle and we can still mark off the points on the Y axis, we still have 04. And on the X axis, we have 30 and 70. But the region that's shaded in for both of these inequalities so far is just from 3-7 for the X values and from for four and up for our Y values. So this is what we're talking about thus far. Now, we just have one more inequality to graph. So let's work on graphing negative two X plus two Y is less than or equal to eight. What we recall from previous lessons that when we're graphing an inequality, the first step is that we're going to set it equal to an equality. So negative two X plus two Y equals eight and now we are going to create an X Y table. We see that our equation is of a line so we can graph the intercepts plugging zero in for X two to graph the Y intercept. Let's do that. We have negative two times no longer X but now zero plus two, Y equals eight, negative two times zero is zero. So we have two Y equals eight, divide both sides by two and we get Y equals four. So we can graph the .04 from the origin. We already have 04. Then that will be one of the points of this line. Now let's graph our X intercept. Plugging zero in for the Y values we have negative two X plus two times not Y but zero equals eight and two times +00. So this is just negative two X equals eight, dividing both sides by negative two, we get X equals negative four. So we can graph the 0.-40 from the origin. We're moving to the left floor units and then neither up nor down for our y value of zero. That's going to be somewhere over here. I'm just going to draw that. Now, imagine a line going through those two points and then continuing through the region that we have, that would look something like this. So now we need to discern if we're talking about the region above this line or below this line. So in order to determine if we're talking about above or below our line, we need to choose test points and the test points don't need to be inside of this region for all three inequalities. It only has to be above and below the particular inequality that we're talking about right now. So I'm going to choose test points of And then another test point above the line can be at 1 100. It'll be far above this on the graph, but clearly it will be above the line. So let's start with 11. We'll see if we're talking about the region below the line. So plugging in one for the X value negative two times one plus two times one for the Y value is less than or equal to eight, negative two times one is negative two plus two times, one is two less than or equal to eight? Negative one, excuse me, negative two plus two equals zero. Is it true that zero is less than or equal to eight? Yes, it is. So we are talking about the region below our line. So this is the graft region that we're talking about. And next, we need to identify the corners of each part of our graft region. So the corners, some of them, we know we know this one, the X value is going to be three and the Y value is going to be four, moving to the right, the next corner here we know that the X value is going to be seven and the Y value is going to be four. But we don't know the corners that intersect with this line of negative two X plus two Y equals eight. But it's not too hard to figure them out because we know they're X values. We know the X value for the one on the left is going to be three. We just don't know the Y value and we know that the X value for the one on the right is going to be seven. We just don't know the Y value. So we can plug the X values in to our equation negative two X plus two, Y equals eight. So let's start off plugging three in for our X value, we have negative two times not X but three plus two Y equals eight. And negative two times three equals negative six plus two, Y equals eight, adding six to both sides. We get two, Y equals 14, dividing both sides by two, we get Y equals seven. So when X equals three, Y equals seven, now let's plug seven in for the X value and figure out the corresponding Y value. So negative to not times X but now times seven plus two, Y equals eight, negative two times seven is negative 14, bringing down plus two, Y equals eight will add 14 to both sides. We have two, Y equals and eight plus 14 is going to equal 22 dividing both sides by two, we get Y equals 11. So when X equals seven, Y equals 11, and now we've identified all four of our corners. The next step is to evaluate the objective function at each corner of the graft region. Now that we finally got this graft. So in order to do that, we're going back to this objective function which is Z equals five X minus Y. So we're looking at the Z value for each of the four corners. Let's start with 34. That means that we are going to plug in three for our X value. So five times not X but three and we're going to plug in four for the Y value. So minus not why but four, While five times 3 is 15, bringing over the -4 and 15 -4 equals 11. Now let's keep going around the next corner we'll look at is 37. So plugging in three for X will be five times three minus plugging in seven for Y. So minus 75 times three is still 15, bringing over the minus 7, 15 minus seven equals eight. Now, looking at the next corner, we'll look at the corner of 7 11. Plugging in seven for the X value will be five times seven And then plugging in 11 for the y value -11 Five times seven is 35, -11 equals 24. Now, the last corner we have is 74 plugging in seven. For the X value, we have five times seven again minus and for the Y value of 45 times seven is still 35 minus 4, 35 minus four is 31. And now we need to identify the maximum value. So looking at these Z values for each of them 11 8, 24 31 the maximum one is 31. So when we have an X value of seven and A Y value of four, we have a maximum value at 31. Alright, we have found our answers. Now, we just need to find the matching answer choice. So let's start off by looking for a graph that has a point at 34, we'll answer choice. A has a point at 23. That one doesn't work in the lower left hand corner of our figure. Answer choice B has a 0.23. Again, that doesn't work. So now let's bring these answers down and we'll go look at C and D and dragging all these Z values down. It'll be a mess for a moment. There's our Z values. Now, let's grab the graph. It'll be a mess for a moment. Okay. So looking at C and D, they have the points at 34377 11 and 74. That's great. Now, let's look at those values for the objective function. Well, where we have three and four, Z equals 11 for answer choice. C that's looking good where we have the 0.37, Z equals eight. That looks good for 11. The equals 24. Great for 74, Z equals 31 with a maximum value of at X equals seven and Y equals four. All right. Answer choice. C works well done. We'll catch you on the next one.

Key Concepts

Graphing Linear Inequalities

Objective Function

Feasible Region and Corner Points

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