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(3-3)+V _ ] tbat is closet (o the (6 paints) Use Lagrange $ miultiplier to fnd & point on Questioni org:n...

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(3-3)+V _ ] tbat is closet (o the (6 paints) Use Lagrange $ miultiplier to fnd & point on Questioni org:n

(3-3)+V _ ] tbat is closet (o the (6 paints) Use Lagrange $ miultiplier to fnd & point on Questioni org:n



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Use the following information. The senior class of Westwood High School wants to paint the entrance hallway floor of their school as shown at the right. CAN'T COPY THE FIGURE Paint costs 20 dollars per gallon. Five gallons of paint covers 2000 square feet. How much will paint cost if the students use four coats of paint?

For this problem we have the value of all the nylon brush is that a paint store Carries is $650. Complete the table and then Foreman equation for this number value problem. So you can see that we have this chart here. The end of the chart is cut off. So the total that was cut off I'm not just right on the side. Here. Total is gonna be the 670 given to us in the word problem. And then we have to figure out the total value so we can see that number times values equal to the total value, so the total value of the one inch brush is too. X times for the total value of the two inch brush is five x or five times X. The total value of the three inch brush is seven times X plus 10 and then thio Foreman equation. For this problem, we're gonna have to add the total values for the one inch, two inch and three inch and make it equal to the 670. So we have two x times for which I'm gonna simplify right now to be eight x. So for our relation, we have a X plus five eggs, plus seven times X plus. 10 is equal to 670. So then we have to combine our like terms. But before we can do that, we have to have to distribute the seven to each of the parts that are inside the parentheses. So we have seven times X and then seven times 10. So we have eight x plus five x, and then we have seven times X, which is seven X and then we have seven times 10 which is 70 is equal to 670. Then we have eight experts. Five X plus seven X is equal to 20 X plus. The 70 is equal to 670 and that is gonna be our equation for this number of value problem.

This problem reads. The value of all the nylon brushes that a paint store carries is $670. Complete the table and then foreman equation for this number value problem. So we have number, times values, sickle to total value, and all of the number and value columns are filled, so we just have to fill out the Total Value Column as well as the total. So starting off, we know that the total is 670. And to find the total value of each nylon brush type, we're gonna have to multiply the number of times the value. So two X times four is equal to eight X, then for the two inch the total values and be five X, and for the three inch, it's gonna be seven times X plus 10. So from here, we're gonna have to create an equation that combines the total values of each brush type. So we have a X plus five X plus seven times X plus ton is equal to 670. So from here we're gonna have to distribute the seven to the X as well as the seven to the 10 So we have a X plus five eggs plus seven times x is seven X and then seven times 10 is 70 is equal to 670. And now I'm going to combine all the light term. So we have eight experts, five X plus seven X, and that's 20 X plus 70 is equal to 670. And from here, I just want to get the 70 to the other side so that the like terms are isolated. And so I'm gonna subtract 70 on each side. So then we're left with 20 X is equal to 600 and then you only have to do one more step to solve for the value of X. So our equation is 20. X is equal to 600.

So we have, again now we're going to get to the ground multipliers. So we're doing what we um what you learned in the chapter. So we have, let's see here, we have um f of X. Y equals two times two X. Y. And then under the constraint X plus Y equals six. So this is a plane cutting through this. And so this would actually be again an easy one to do without the garage multiplayer, but it's not hard to do with the grunge multiplayer. So we just we take this function on augmented with statistical, make this uh function and find a function where such that this is equivalent to this but equal to zero and that's just X plus Y minus six. So we take this function and I'll add to it the grounds multiplier by the function we made from here. And so we need to take the essentially the gradient or the, you know, the partials with respect to X, X, Y and lambda. And so I just denoted that with a subscript, you know, kind of a subscript of a list here. And so then I have a list of values here. So when I take the door of the respect to X, um I get this and evaluated at, you know, the critical point because I'm gonna set it equal to zero, take through respect of why we do this take to respect the land that we just get our constraint back. And that will always be the case just because we, you know, added lambda time with this Um and set that equal to zero and we wind up with with X equals 23 Y equals three and lambert is minus six. So I didn't plug it back in, so we get three, three go 9 18. And so that's um it's hard to, I mean it would be easy to see if we did the substitution. That this is actually a maximum. It's hard to know here unless we went to a different uh unless we went to a different did take put more derivatives to higher derivatives to figure out if that was a minimum actually. But there's only, There's only one critical point. So we can easily just kind of plug in values here to see, you know, keeping this constraint and see, you know, do we ever get a value, Does it look like this is the minimum or maximum? So it's not hard to check and this is in fact it was a later, it was a maximum I think I'd say a minimum before it's a maximum. This is a baxam. Um Next one we have six xy subject, two by minus X equals six. So this is again a plane cutting through this and it would be easy to solve without the grand multi players, but we use it so we take this function and augment it with the functional equivalent of this, that's equal to zero. Um and that's this by -X -6. And take the partials with respect to X, Y and lambda. And we get these three functions And set them all equal to zero and then solve them simultaneously. So we have three equations, three unknowns and we get the extremely point is that X equals minus three bicycles three. And lambda is 18. Now we plug in, I didn't plug back in so we have minus three, three. So that is, let's see here, we've got nine 54 -54. Yeah -54. So that is a there's also a minimum. Yeah, yeah, that's him. So can we can plug in can testing take test points that, you know, we could just plug this back into here and you can see that. Yeah. Let's see. Is it a minimum by equals? Yeah. Because this will give us a problem. That's upwards now, let's see here. The next one we have f of x equals X squared plus Y squared subject to three X plus. For a while it was 25. This is still just a plain. And you could obviously solve for X or Y and substitute back into here. But using the grounds multipliers, we take this function um make a function of the out of this equation such that the function equals zero. And we get this added to this but multiplied by lambda. And we get this expression here and that we need to minimize over X, Y and lambda. So we need to take partial with respect to X, Y and lambda. And so we get these three functions here And we need to see here, we need to set that to zero evaluated at the critical point. Set it equal to zero salt for three equations, three unknowns. And we get X equals three white four and lambda equals minus two. Now, I mean obviously you don't need to solve for landed here. And I've never actually, I've never actually seen anybody any argument for if this is at all useful. Does this, I've always wondered does this, does this tell you anything? And I can't ever think about what it might actually tell you. Um It just happens that you get it, but it's not any information that you might use in any way that I can think of. Uh So we plugged back in which I forgot to do for, so we get nine plus 16 equals 25. So that's uhh and obviously it's kind of interesting because this winds up being just X. Squared plus Y squared equals 25. So it happens it just happens to be that point. And this is a let's see here this is a minimum. Yeah you can figure that out you know fairly easily. But again you'd have to take the second partials to really prove that. But you could actually just figure it out because there's only one extreme appoint. So if you plug in any other number that satisfies this constraint, plug it into here. Numbers that satisfy this and plug into here you're going to get something that's That's bigger than 25. Which means that this had to be a minimum because there's only one extreme appoint.

So we have again now we're gonna get to the ground multipliers. So we're doing what we um what you learned in the chapter. So we have, let's see here, we have um f of X. Y equals two times two X. Y. And then under the constraint X plus Y equals six. So this is a plane cutting through this. And so this would actually be again an easy one to do without the garage multiplayer, but it's not hard to do with the Grunge multiplayer. So we just we take this function on augmented with statistical, make this uh function and find a function where such that this is equivalent to this but equal to zero and that's just X plus y minus six. So we take this function and I'll add to it the grounds multiplier by the function we made from here. And so we need to take the essentially the gradient or the, you know, the partials with respect to X, X, Y and lambda. And so I just denoted that with a subscript, you know, kind of a subscript of a list here. And so then I have a list of values here. So when I take the door of the respect to X, um I get this and evaluated at, you know, the critical point because I'm gonna set it equal to zero, take the respect of why we do this take to respect the land that we just get our constraint back. And that will always be the case just because we, you know, added lambda time with this Um and set that equal to zero and we wind up with with X equals 23, Y equals three and lambert is minus six. So I didn't plug it back in, so we get three, three go 9 18. And so that's um it's hard to, I mean it would be easy to see if we did the substitution that this is actually a maximum. It's hard to know here unless we went to a different uh unless we went to a different did take put more derivatives to higher derivatives to figure out if that was a minimum actually. But there's only There's only one critical point. So we can easily just kind of plug in values here to see, you know, keeping this constraint and see, you know, do we ever get a value? Does it look like this is the minimum or maximum? So it's not hard to check and this is in fact it was a later it was a maximum I think I'd say a minimum before it's a maximum. This is a baxam. Um Next one we have six. Xy subject two by minus X equals six. So this is again a plane cutting through this and it would be easy to solve without the grand multi players, but we use it so we take this function and augment it with the functional equivalent of this, that's equal to zero. Um and that's this by -X -6. And take the partials with respect to X, Y and lambda. And we get these three functions And set them all equal to zero and then solve them simultaneously. So we have three equations, three unknowns. And we get the extremely point is that X equals minus three bicycles three and lambda is 18. Now we plug in, I didn't plug back in so we have minus three, three. So that is let's see here, we've got nine 54 -54. Yeah -54. So that is a there's also a minimum. Yeah, yeah, that's him. So can we can plug in can testing take test points that, you know, we could just plug this back into here and you can see that. Yeah. Let's see. Is it a minimum? Why equals? Yeah. Because this will be give us a problem. That's upwards now, let's see here. The next one we have f of x equals X squared plus Y squared subject to three X plus. For a while it was 25. This is still just a plain. And you could obviously solve for X or Y and substitute back into here. But using the grounds multipliers, we take this function um make a function of the out of this equation such that the function equals zero. And we get this added to this put multiplied by lambda. And we get this expression here and that we need to minimize over X, Y and lambda. So we need to take partial with respect to X, Y and lambda. And so we get these three functions here And we need to see here, we need to set that to zero evaluated at the critical point. Set it equal to zero salt for three equations, three unknowns. And we get X equals three white four and lambda equals minus two. Now, I mean obviously you don't need to solve for landed here. And I've never actually, I've never actually seen anybody any argument for if this is at all useful. Does this, I've always wondered does this, does this tell you anything? And I can't ever think about what it might actually tell you. Um It just happens that you get it, but it's not any information that you might use in any way that I can think of. Uh So we plugged back in which I forgot to do for, so we get nine plus 16 equals 25. So that's uhh and obviously it's kind of interesting because this winds up being just X. Squared plus Y squared equals 25. So it happens it just happens to be that point. And this is a let's see here this is a minimum. Yeah you can figure that out you know fairly easily. But again you'd have to take the second partials to really prove that. But you could actually just figure it out because there's only one extreme appoint. So if you plug in any other number that satisfies this constraint, plug it into here. Numbers that satisfy this and plug into here you're going to get something that's That's bigger than 25. Which means that this had to be a minimum because there's only one extreme appoint.


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