In Exercises 57–59, graph the region determined by the constra...

Question

In Exercises 57–59, graph the region determined by the constraints. Then find the maximum value of the given objective function, subject to the constraints. This is a piecewise function. Refer to the textbook.

Accepted Answer

Welcome back. I'm so glad you're here. We've got this function the equals six X plus 13 Y. And it's got five constraints here will label them A. B, C and D and E. Because five gets us to E. And we are going to graph those constraints. Let's start with D. And E. D says that why is greater than or equal to zero with its anything up here above the X axis? E says that X is greater than or equal to zero. That's anything over here. To the right of the Y axis. So where did those two constraints come together? The first quadrant here. Alright. We're only dealing with the first quadrant. Great. We've got two of our constraints figured out it's going to take us a little bit longer to get the other three. So constraint a will work on that 1. 1 says that two X plus three Y is greater than or equal to nine. Okay, let's treat that as a line first, two, X plus three, Y equals nine. Let's figure out the X and Y intercepts plug zero in for X. Two times zero plus three Y equals nine. We've got three, Y equals nine, divide all sides by three, Y equals three. So that is our Y intercept zero on the X axis up about three on the Y axis. Now let's figure out the X intercept zero in for Y. We've got two X plus three times zero equals nine. Gives us two X equals nine viable sides by two and X equals nine halfs or 4.5. So we look on our X axis go over about 4.5. And let's call that about here and we connect our two dots, connect our points, draw a line label that a Now, which region are we talking about when we've got two X plus three Y is greater than or equal to nine. What is it this region up here? To the right or is it this region down here? To the left? While we can figure that out by entering a test point into that inequality. And the easy test point to use is +00. So let's do that two times zero plus three times zero is greater than or equal to nine. Is that right? Is zero plus zero is zero greater than equal to nine. No, so our test point failed but that's okay. It just tells us that we're talking about this region up here specifically within the first quadrant because that's what we learned from D and E. Right. So we got information from our failure. Great. Now let's figure out constraint B. B tells us that three X plus two Y is going to be less than or equal to 13. Cool. Will treat that as a line first, three X plus two, Y equals 13. And we are going to plot our X and Y intercepts will put zero in for X first. So three times zero has nothing plus two. Y equals 13. So two Y equals 13. The bible sides by two. And we've got Y equals 13 halves which is the same as 6.5. So we're gonna look on our graph or 6.5 on the Y axis. Nothing on the X axis. That can be here. Now let's get our X intercept awesome. Well plug zero in for Y. For this 13 X plus two times zero. That's nothing equals 13. So we've got three X equals 13 bible sides by three. We've got X equals 3rd Or four in a third. Well that's interesting to note because four and a third is right before the 4.5 we had on the X axis for a value. So we are drawing this line very close to but right in front of are a weird intersects with the X axis. So let's connect these two points ish. And label that be. It's just a rough rough sketch. Now. Are we talking about this region up here to the right? When we say three X plus two, Y is less than or equal to 13. Are we talking about this region down here? To the left and again we can use our test point to figure that out. +00 is our friend. So three times zero plus two times zero is less than or equal to 13. Ok. Again the left hand side is zero. So is it true that zero is less than or equal to 13? It is that one is true. So that includes our test point this time but it doesn't go all the way to our test point because it really only goes up to where this a line is. So it includes the shaded region of both. So now we're just looking at that triangle and we've figured out constraints A. B, D. And E. Let's just do see. So C tells us that why is less than or equal to X. Okay let's treat that as a line. Y equals X. Alright. Are X and Y intercepts? Okay, zero in for X. Well if zero X ray, Y is also zero. So 00 that's the origin. And we plot one more points. We need two points for a line. If we put a one in for Y. That will be a one R. X. Value to. So 11 to the right. One for our X. And up one for ry can be about here. And now we draw a line connecting our two points and that can look like this label at sea. And now are we talking about this region down to the right or this region up to the left. Well we'll pick a test point but unfortunately our friend 00 is part of this line. So we can't use that. Let's pick here. Let's pick two for X. Value. Zero for our Y value. Just drawing it as a line so that we can really see where and I can erase it easily without erasing our whole graph. So we put two in for X is greater than or equal to zero for Y. Is that true? Yes, it is zero is less than or equal to two. So it's this region but it's not that whole region. It's only the part that intersects with what we've already got shaded in. So it's just this region here. Great. We have got a big chunk of this one done. That's the region that's constrained by these five constraints. And now we've got to figure out what the values are at the three points where they intersect. And in order to do that, I need more room. So I'm going to move this down so that we don't forget it. We've got the three equations for our three lines and to find out the values of X and Y where they intersect. Let's start with A. And B intersecting at this point. So we're going to use the lines equations. So that will be two, X plus three, Y equals nine for A and three X plus two, Y equals 13 for B. Now we've got a system of equations and we can solve for X and Y. Let's get our X values to cancel out. So we're going to multiply that first row by three in the second row by negative two. So our new first row will be six, X plus nine Y equals 27. And our new second row will be negative six X minus four, Y Equals -26. And we can now just add those. So our X's cancel out nine Y minus four Y. Gives us five Y equals 27 plus negative 26 gives us one. So we've got Why equals 1/5. And we'll start labeling that up here. 1/5 is the same as 0.2. We'll use that 0.2 value on our graph. Great. Now we'll plug the Y value in and I'll use 1/5 up here. So two X plus three times 1/ equals nine. So it's two X plus 3/5 equals nine. Okay, subtract 3/5 from both sides. We've got two X equals 42. 5th. Yes I did this beforehand, divide both sides by two and we've got our X value which is 21 5th, C. Or 4.2. Great. We've got our X and Y values. I'll label that up here. So our X value is 4.2. Great. one Down 2 more to go. Now let's work on the intersection of A. And C. Right here. So our a value or a equation is good to go but we've got to do a little bit of work with our c equation. Let's scoot this Y over. So it's on the same side as our X. And remember we're dealing with we were dealing with Y equals X now so subtract Y from both sides and it's zero equals X minus Y. Or in other words X minus Y equals zero. And we can work with that equation much easier. Oops. So now we've got two X plus three, Y equals nine and x minus Y equals zero. We want to add these two together but we want let's get our wise to cancel out. So let's multiply the second row by three. So two X plus three times X. That'll be two, X plus three. X. Gives us five X wise cancel out equals nine plus zero times three, that's zero. So nine, five X equals nine survival sides by five and X equals 9/5 C. Or 1.8. And now we're going to use a cool trick here, we know that Y equals X on this line, so why equals 1.8. Cool. We'll label that up here 1. And 1.8. And now let's get this last point where B and C intersect, go back down to our working space here. So be the equation is three X plus two, Y equals 13. We are R. C. Equation from the last one. We've got X minus Y equals zero. We want to add these together. Let's get our wise to cancel out again. Let's multiply the second row by two. So three X plus two times X. Is two X. So that's five X. Again are wise cancel out equals 13 plus zero times to zero. So 13 divide both sides by five and we've got X equals 13 5th so or 2.6. And the cool thing again, we know that our Y value is equal to our X. Value. So why equals 2.6. We'll head back up and label this on our graph. 2.6. 2.6. You guys we're almost there. This has been a journey because now we get to go back to that objective function where Z Equals six x plus 13 y. And we're going to plug in our X. And Y values. We're going to get three answers here. So let's use our 1st 30.4 point 20.2 in black. So Z equals six times the X value is 4.2 Plus 13 times the y values. 0.2. And my calculator yields the answer of 27.8. Okay, that's one value. Now let's do the purple the 1.81 point eight. So Z equals six times 1. plus 13 times 1.8. And the calculator lets us know that that is 0.2. And the final one we'll use 2.62 point six. Z equals six times 2.6 plus times 2.6. And that gives us the answer of 49.4. And we look at these and we say which one is the biggest. Well that's the third one. We look at our answer choices and that's answer choice. C Well done, we'll catch you on the next one.

In Exercises 57–59, graph the region determined by the constraints. Then find the maximum value of the given objective function, subject to the constraints. This is a piecewise function. Refer to the textbook.

Key Concepts

Graphing Constraints

Objective Function

Piecewise Functions

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