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-/2 pointsTanFin12 4.3.015.CMIUse the method of this section to solve the linear programming problem Minimize C = 2x 3y + 82 subject to ~X + 2y 2 < 20 2y 2z <...

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-/2 pointsTanFin12 4.3.015.CMIUse the method of this section to solve the linear programming problem Minimize C = 2x 3y + 82 subject to ~X + 2y 2 < 20 2y 2z < 10 2x + 4y 3z < 12 x 2 0,Y 2 0,2 2 0The minimum is C =(x, Y, 2) =Need Help?Acad lSubmit AnswerPractice Another Version

-/2 points TanFin12 4.3.015.CMI Use the method of this section to solve the linear programming problem Minimize C = 2x 3y + 82 subject to ~X + 2y 2 < 20 2y 2z < 10 2x + 4y 3z < 12 x 2 0,Y 2 0,2 2 0 The minimum is C = (x, Y, 2) = Need Help? Acad l Submit Answer Practice Another Version



Answers

Use the simplex method to solve.
Minimize $$w=3 y_{1}+2 y_{2}$$
subject to: $$\begin{aligned} y_{1}+y_{2} & \geq 8 \\ 2 y_{1}+y_{2} & \geq 12 \end{aligned}$$
with $$y_{1} \geq 0, \quad y_{2} \geq 0$$

Okay. This question gives us this objective function to optimize subject to these linear constraints. So plotting this region and Daz Mose, we come up with this graph and since we have less than signs everywhere, that means that the region we're worried about is this part right here in the first quadrant. So what this means is that our vergis ese are here, Here, here, in here. And this defines a regent. So this vertex rate here iss 50 This vertex right here is the origin. This Vertex right here is 36 and this Vertex is 0 10 So now that we found our four Vergis ese, we just need to test them into our objective function. So Z of 00 is just zero plus zero. Which zero then see of 0 10 is equal to zero plus 10 which is equal to 10. Then Z of 50 is equal to two times five plus zero, which is also equal to 10. Then Z of 36 is equal to two times three plus six, which is equal to 12. So what we can see here is that our minimum value is zero and our maximum value his 12th. So now that we have this information, we can just fill in that our maximum value is 12 at the Vertex 36 and our minimum zero at the origin. And just to note here we see that 0 10 and 50 for deceives actually have the same value. So that means that anything on this lavender colored line in our feasible region, we'll have the value of 10 also, but it doesn't really matter. Since it's not a maximum in, it's just something worth noting.

In this case we want to maximize the product the two numbers subject to the constraint that the some of those numbers equals 10. So that X plus y minus 10 is zero. So we have to surface and then um we have this this is a line then. So we're looking at the surface and if we look down from the top, you know along some line were traveling along and we're trying to find um the maximum. Now in this problem it would probably be easy to substitute sulfur Y and subsequent acts and substitute into here. But we can use lagrange multipliers by augmenting dysfunction with our constraint taking the partial with respect to X. The partial perspective why? In the partial perspective lambda And setting all those 30 with three questions and three announcing the linear equations. So there fairly easy to solve and we get X equals five and Y equals five and lambda is minus five. Um If you care what that is And then if we plug that back into here, we can see clearly that these some 10 and you get plugged back in here and you get 25. So the maximum of any two numbers, some 2 10, author. The matrix of the product of any two numbers, some 2 10 is 25. No, now we're asked to find the product maximum. The product of any two numbers were actually the minimum of a product of any two numbers that um some to whose differences minus 10, I think that's a minimum. Yeah, So we plug in. Uh huh. Yeah, it'll be if you solve for X and plug into here, we're gonna get a minimum if you solve for X and plug into here. Again, we're going to get a minimum. Okay, now we have basically two constraints that we have to look at because the difference that is equal to 10, so we can either have x minus y or y minus X. They didn't really say which order. So we can basically look at both of them and it's basically the same problem. Um So we augment our function with our constraint and I just use this constraint for here. You can easily use this one instead you just change the sign on um Change the sign on excess and wise. So take the gradient or to take to foster respect to X and wide and lambda except um equals zero and we get X equals five, Y equals minus five and landed equals five. Now if we change the change X. Y, change the sine of X and Y. We're just going to change the sign of the results. Yeah. Um So we get the other solution is X equals minus five. Y. It was five. So again the difference, you know this minus this is 10. This minus this is 10. And so in either case the product is minus 25. Mhm. And that's a minimum. Yes. Now we have um going into three D. We have a this is basically a contours of this are spheres. So we have basically a sequence of nested spheres basically continually nested, This is a plane in three D. So basically we're looking at, you know, we have this plane somewhere and we're looking for these spheres and we're looking for the maximum sphere that touches the plane basically. And you know, in two D. You can think about, you know, you have a line and a bunch of circles and you're looking for the maximum where the circle touches the line. And so that's kind of the geometric interpretation is mathematically we can just take this function augmented with lambda towns that constraint. And then we take the partial with respect to X, Y. Z. And lambda. So we wind up with four equations here for unknowns, X, Y. Z. And lambda set. So we have four equations. And for unknowns there homogeneous, you know, they're not homogeneous because we have constant. Over here, They were homogeneous and everything would be zero. But we can solve them, they're living here, so we solve them and we get X equals minus four, Y equals two, Z equals minus six. And landed for completeness is for And so we plugged that into here into this and we get 16 plus four plus 36. That's 40. So we get 56. So I sphere with a radius of square with a 56. We should just touch this plan here.

In this case we want to maximize the product the two numbers subject to the constraint that the some of those numbers equals 10. So that X plus y minus 10 is zero. So we have to surface and then um we have this this is a line then. So we're looking at the surface and if we look down from the top, you know along some line were traveling along and we're trying to find um the maximum. Now in this problem it would probably be easy to substitute sulfur Y and subsequent acts and substitute into here. But we can use lagrange multipliers by augmenting dysfunction with our constraint taking the partial with respect to X. The partial perspective why? In the partial perspective lambda And setting all those 30 with three questions and three announcing the linear equations. So there fairly easy to solve and we get X equals five and Y equals five and lambda is minus five. Um If you care what that is And then if we plug that back into here, we can see clearly that these some 10 and you get plugged back in here and you get 25. So the maximum of any two numbers, some 2 10, author. The matrix of the product of any two numbers, some 2 10 is 25. No, now we're asked to find the product maximum. The product of any two numbers were actually the minimum of a product of any two numbers that um some to whose differences minus 10, I think that's a minimum. Yeah, So we plug in. Uh huh. Yeah, it'll be if you solve for X and plug into here, we're gonna get a minimum if you solve for X and plug into here. Again, we're going to get a minimum. Okay, now we have basically two constraints that we have to look at because the difference that is equal to 10, so we can either have x minus y or y minus X. They didn't really say which order. So we can basically look at both of them and it's basically the same problem. Um So we augment our function with our constraint and I just use this constraint for here. You can easily use this one instead you just change the sign on um Change the sign on excess and wise. So take the gradient or to take to foster respect to X and wide and lambda except um equals zero and we get X equals five, Y equals minus five and landed equals five. Now if we change the change X. Y, change the sine of X and Y. We're just going to change the sign of the results. Yeah. Um So we get the other solution is X equals minus five. Y. It was five. So again the difference, you know this minus this is 10. This minus this is 10. And so in either case the product is minus 25. Mhm. And that's a minimum. Yes. Now we have um going into three D. We have a this is basically a contours of this are spheres. So we have basically a sequence of nested spheres basically continually nested, This is a plane in three D. So basically we're looking at, you know, we have this plane somewhere and we're looking for these spheres and we're looking for the maximum sphere that touches the plane basically. And you know, in two D. You can think about, you know, you have a line and a bunch of circles and you're looking for the maximum where the circle touches the line. And so that's kind of the geometric interpretation is mathematically we can just take this function augmented with lambda towns that constraint. And then we take the partial with respect to X, Y. Z. And lambda. So we wind up with four equations here for unknowns, X, Y. Z. And lambda set. So we have four equations. And for unknowns there homogeneous, you know, they're not homogeneous because we have constant. Over here, They were homogeneous and everything would be zero. But we can solve them, they're living here, so we solve them and we get X equals minus four, Y equals two, Z equals minus six. And landed for completeness is for And so we plugged that into here into this and we get 16 plus four plus 36. That's 40. So we get 56. So I sphere with a radius of square with a 56. We should just touch this plan here.

In order to find the min and Max of this objective function, we should first sketch the constraint graph. So in order to do that, we can take each inequality in front of X and Y intercepts. So if we start with this 1st 1 and if we plug in X equals zero, then we got four. White is law center and golden 20. And if we divide both sides before I get why it's less cynical of five, and that gives us the 0.0 comma five right there. If we plug in Y equals zero, we get out. X is less than or equal to 20 which gives us the point 20 comma zero, which is right here. We can connect those two points. Morris get our first line. So if you look at the next inequality, we can do the same thing. Put an X equals zero when they get out. Why was less than an angle to 18 which gives us the point zero comma 18 which is somewhere right around here. And then if we plug in Y equals zero, we get the same thing for X X is less than an equal to 18 gives us the point. 18 karma. Zero. It's right around here. And then we can connect those two points. Get our second line. Is that what's my Greek? Okay. For our last inequality, you plug in X equals zero and we get to wise last article to 21. We divide both sides by two. We get wise, listen or equal to 10.5 and then if we plug and y equals zero, we get the same thing for axe, which is X is listening 10.5 and those two give us a 20.0 comma 10.5 and 10.5 Common zero. Just ride around here. Really? Connect those two points. Get our final line. Now we want to try to find which area to shade. So we noticed in the given constraints that why x Mr Creativity go to zero Come trains us to the positive axes here. With her first inequality, we noticed that X must be lost. Undergo 20 Here we noticed that X must be less than angle to 18 And here X must be less cynical to 10.5, which means putting this all together X must be listener called a 10.5. So we know that X is gonna be somewhere too. The left of this. This line. So now we look at why so in the 1st 1? Why must feel cynical to five. The 2nd 1 is 18 or a journal in his 10.5. So putting all those together, why must the last honorable of five, which must mean that Why must be under this line. It's an X must be to the left of this line. We can trade in this area that from that we can figure out what points we want to plug into the objective function. So that means we want these points. We want zero comma zero. We want 10.5 comma. Zero want, dear. Oh, come at five. And then we need to figure out this what? This last point ISS. So in order to do that, we can set up a system offline your questions using these two ones. So it looks like this line front comes from the Inequality X plus four wives gold 20. The other line comes from this third inequality, so we can set up a system where we try to cancel out. Don't say we could try to cancel out the X variable. So then we just figure out what wise In order to do that, we can multiply this first inequality by two and subtracted 3rd 1 so that we get Seo for X. So let's do that. We take the 1st 1 We got two x plus two times four is eight. Why listen, every pulled into turns 20? It's 40. We subtract the whole the entire third inequality two x close to Why Just listen and call the 21. Then we get to explain to x zero. If I'm honest to I six, why 40? Minus 21? It's 19. That gives us why is less than or equal to 19 over six. Um, okay, Now you put that into one of these inequalities. Large jacks. So X plus four times 19 over six is less than any pull at 20. So if we simplify this part, we can simplify this to two times 19 over three. Okay, And then if we want to find that out, we get 38 over three. So that gives us X is listening and will do 20 minus 38 over three. And if we kind of some fire, they're speaking at 60 over three, minus 38 over three, which gives us 22 or three. And that gets us our final point. 20 over three karma 19 over six. Basically, we need to plug in all of these points into, um, the objective function in order to figure out the max. So if we take, the 1st 1 was pregnant until this objective function can't see equals zero. Okay, If we take the next point, get C equals two times 10.5, which is 21 close zero, and we look at the 3rd 1 C equals zero plus four times five is 20. And then our last one we get C equals 22 over three times two is 44 over three and then plus four times 19 over six. And we actually did that exact same calculation right here and then got 38 over three, not equal to 82 over three, which is around, um, 27.3 approximately. So that means this is our maximum. So we have a maximum of 82 over three at this point. Right here, Mr Max. And then we have our minimum zero at 0.0


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