For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Further Examples with Extrema

^{2}on the region D = { (x,y)| − 1x ≤ 1, − 1 ≤ y ≤ 1} .

- The region D is closed and bounded and f is continuous on D, so there exists a maximum and minimum value on D. We first look for critical points on that region.
- So [df/dx] = 0 implies 2 = 0 which is not true. Also [df/dy] = − 4y = 0 implies y = 0. Then our critical points occurt anywhere at f(x,0) = 2x, that is, along our x - axis. Note that this is an increasing function on the xz - plane.
- We now have f( − 1,0) and f(1,0) as extrema. Since f is increasing we must check the endpoints of our region D for extrema as well.
- Now f( − 1,1) = − 4 and f( − 1, − 1) = − 4. We see that both of these values are less than f( − 1,0) = − 2.

- The region D is closed and bounded and f is continuous on D, so there exists a maximum and minimum value on D. We first look for critical points on that region.
- So [df/dx] = [1/(2√x )] + y also [df/dy] = − 1 + x = 0 implies x = 1. Hence [1/(2√1 )] + y = 0 implies y = − [1/2] and so our critical value is at ( 1, − [1/2] ). Note that ( 1, − [1/2] ) ∉ D.
- To check for extrema on our boundary we observe the four cases: f(0,y), f(2,y), f(x,0) and f(x,1).
- For f(0,y) we have z = − y whose maximum value on D occurs at y = 0 and minimum value occurs at y = 1. Our possible extrema are (0,0,0) and (0,1, − 1).
- For f(2,y) we have z = √2 − y + 2y = √2 + y an increasing function on D. The minimum value occurs at y = 0 and maximum value at y = 1. Our possible extrema are (2,0,√2 ) and (2,1,√2 + 1).
- For f(x,0) we have z = √x which is an increasing function. On D the minimum value occurst at x = 0 and the maximum value at x = 2. Our possible extrema are (0,0,0) and (2,0,√2 ).
- For f(x,1) we have z = √x − 1 + x. Note that √x − 1 and x are increasing functions and so z = √x − 1 + x is also increasing. Our possible extrema are at (0,1, − 1) and (2,1,√2 + 1).

^{2}− 4x

^{2}√y on the region D = { (x,y)| − 1 ≤ x ≤ 1,0 ≤ y ≤ 2} .

- The region D is closed and bounded and f is continuous on D, so there exists a maximum and minimum value on D. We first look for critical points on that region.
- So [df/dx] = 0 implies − 8x√y = 0 with solutions x = 0 or y = 0. Also [df/dy] = − 6y − [(2x
^{2})/(√y )], if x = 0 we obtain y = 0. A critical value is (0,0,0). - Since y0 we find that [df/dy] = − 6y − [(2x
^{2})/(√y )] = 0 yields − 3 = ( [x/(y^{3 \mathord/ phantom 3 4/ 4})] )^{2}which is not possible. - But recall that y = 0 is undefined at [df/dy], so our critical values are the points (x,0,0). Note that at x = 0 we have (0,0,0).
- Thus far our analysis has given us a possible minimums at (x,0,0).
- To check for extrema on our boundary we observe the four cases: f( − 1,y), f(1,y), f(x,0) and f(x,2).
- For f( − 1,y) we have z = 3y
^{2}− 4√y which is a part of a parabola on D. We find the minimum value of the parabola (vertex) to be − 3^{2 \mathord/ phantom 2 3 3}, so our possible extrema becomes ( − 1,3^{ − 2 \mathord/ phantom 2 3 3}, − 3^{2 \mathord/ phantom 2 3 3}). - Similarly f(1,y) results in z = 3y
^{2}− 4√y . We obtain the extrema ( 1,3^{ − 2 \mathord/ phantom 2 3 3}, − 3^{2 \mathord/ phantom 2 3 3}). - For f(x,0) we have already obserbed that our extrema is (x,0,0).
- For f(x,2) we have z = 12 − 4x
^{2}√2 which is a part of a parabola on D. We find the minimum value of the parabola (vertex) to be 12, a previously discovered possible extrema ( 0,2,12 ).

^{ − 2 \mathord/ phantom 2 3 3}, − 3

^{2 \mathord/ phantom 2 3 3}), ( 1,3

^{ − 2 \mathord/ phantom 2 3 3}, − 3

^{2 \mathord/ phantom 2 3 3}). Hence our minimum values are ( − 1,3

^{ − 2 \mathord/ phantom 2 3 3}, − 3

^{2 \mathord/ phantom 2 3 3}) and ( 1,3

^{ − 2 \mathord/ phantom 2 3 3}, − 3

^{2 \mathord/ phantom 2 3 3}) while our maximum value is (0,2,12).

^{2}− 4y

^{2}+ 3x − 2y − 1 on the triangular region D enclosed by the vertices (0,0), (1,0) and (0,1).

- The region D is closed and bounded and f is continuous on D, so there exists a maximum and minimum value on D.
- We first look for critical points on that region, [df/dx] = 10x + 3 and [df/dy] = − 8y − 2.
- So [df/dx] = 0 gives x = − [3/10] and [df/dy] = 0 gives y = − [1/4]. The critical value is f( − [3/10], − [1/4], ) but note that ( − [3/10], − [1/4], ) ∉ D.
- To check for extrema on our boundary we observe the three cases: f(0,y), f(x,0) and f(x, − x + 1).
- For f(0,y) we have z = − 4y
^{2}− 2y − 1 which is a part of a parabola on D. The function is decreasing, hence it has possible extrema at (0,0, − 1) and (0,1, − 7). - For f(x,0) we have z = 5x
^{2}+ 3x − 1 which is a part of a parabola on D. The function is increasing, hence it has possible extrema at (0,0, − 1) and (1,0,7). - For f(x, − x + 1) we have z = 9x
^{3}− 3x − 3 which is a part of a parabola on D. Note that the vertex of this parabola is at ( [1/6],[5/6], − [13/4] ) hence it is a possible extrema. Another other extrema occurs at (1,0,3).

^{2})/4] − [(y

^{2})/9] on the triangular region D bounded by the parabola y = x

^{2}and the line y = 1.

- The region D is closed and bounded and f is continuous on D, so there exists a maximum and minimum value on D.
- We first look for critical points on that region, [df/dx] = [x/4] and [df/dy] = − [2y/9].
- So [df/dx] = 0 gives x = 0 and [df/dy] = 0 gives y = 0. The critical value is f(0,0) = 0, a possible extrema is (0,0,0).
- To check for extrema on our boundary we observe the two cases: f(x,x
^{2}) and f(x,1). - For f(x,x
^{2}) we have z = [(x^{2})/4] − [(x^{4})/9] which has extrema at x = ±[3/(2√2 )], but such x values are outside D. - In the interval − 1 ≤ x < 0 the function is decreasing while on the intervals 0 > x1 the function is increasing. Aside (0,0,0) we have the possible extrema ( − 1,1,[5/36] ) and ( 1,1,[5/36] ).
- For f(x,1) we have z = [(x
^{2})/4] − [1/9] which has an extrema at x = 0. The function is a parabola decreasing from − 1 ≤ x < 0 and increasing from 0 > x1. The only new extrema we obtain is ( 0,1, − [1/9] ).

^{x}+ y

^{2}.

- Note that there is no guarantee that there exists a maximum or minimum value. We can begin by finding our critical points.
- For [df/dx] = e
^{x}= 0 we have no possible values for x. For [df/dy] = 2y = 0 we have y = 0. Our critical points occur at (x,0,e^{x}).

^{x}is an increasing function with no extrema. Hence f(x,y) = e

^{x}+ y

^{2}has no maximum or minimum values.

^{2}+ 5y.

- Note that there is no guarantee that there exists a maximum or minimum value. We can begin by finding our critical points.
- For [df/dx] = 2x − 2 = 0 we have x = 1. For [df/dy] = 5 = 0 we have we have no possible values for y. Our critical points occur at (0,y,1 + 5y).

^{2}+ 5y has no maximum or minimum values.

^{2})] + y

^{2}.

- Note that there is no guarantee that there exists a maximum or minimum value. We can begin by finding our critical points.
- For [df/dx] = − [2/(x
^{3})] = 0 we have x undefined at 0. For [df/dy] = 2y = 0 we have y = 0. Our critical points occur at ( x,0,[1/(x^{2})] ).

^{2})] is an increasing function from − ∞ ≤ x < 0 and a decreasing function from 0 > x >∞. Hence f(x,y) = [1/(x

^{2})] + y

^{2}has no maximum or minimum values.

- For [df/dx] = 2x − y − 1 = 0 and for [df/dy] = − x = 0 or x = 0. Then 2(0) − y − 1 = 0 yields y = − 1. Our critical point is at (0, − 1,0).

^{2}+ y

^{2}+ [1/(x

^{2}+ y

^{2})] on the region D = { (x,y)|x

^{2}+ y

^{2}≤ 1}.

- So [df/dx] = 2x − [2x/(x
^{2}+ y^{2})] = 0 which results in x^{2}+ y^{2}= 1. Also [df/dy] = 2y − `[2y/(x^{2}+ y^{2})] = 0 results in x^{2}+ y^{2}= 1. Our extrema are the points satisfying x^{2}+ y^{2}= 1. - Note that these points are the boundary of D, so our extrema must occur at x
^{2}+ y^{2}= 1. - We can use the parametric representation for a circle to find the extrema at the boundary. Set C(t) = (cos(t),sin(t)).
- Then f(C(t)) = ( cos(t) )
^{2}+ ( sin(t) )^{2}+ [1/(( cos(t) )^{2}+ ( sin(t) )^{2})] = 2. - Note that no matter what value of t we input our output is always 2. Hence the points satisfying x
^{2}+ y^{2}= 1 are either a minimum or a maximum. - Now, if x < 1 and y < 1 then x
^{2}+ y^{2}< 1 but [1/(x^{2}+ y^{2})] > 1. So f(x,y) = x^{2}+ y^{2}+ [1/(x^{2}+ y^{2})] > 2 for nonboundary points (x,y) ∈ D.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Further Examples with Extrema

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Further Example with Extrema 1:02
- Example 1: Max and Min Values of f on the Square
- Example 2: Find the Extreme for f(x,y) = x² + 2y² - x
- Example 3: Max and Min Value of f(x,y) = (x²+ y²)⁻¹ in the Region (x -2)²+ y² ≤ 1

### Multivariable Calculus

### Transcription: Further Examples with Extrema

*Hello and welcome back to educator.com and to multi-variable calculus.*0000

*Today we are going to continue our discussion of maximum, minimum.*0004

*We are actually just going to do example problems. The last lesson, we finished off with the main theorem of closed and bounded sets and the existence of absolute max's and absolute min's on that particular closed and bounded set.*0009

*Now, we are going to just going to do some examples to get an idea of the different things that can come up when we are dealing with a particular function.*0022

*Really, this lesson is just about getting comfortable with all of the different tools that we have to use - all that... everything that we know, whatever it is that we can bring to solve a particular problem regarding a maximum and minimum.*0030

*Remembering all along that really what we are trying to find is those points where the function obtains a maximum, and obtains a minimum.*0045

*Ultimately this is really just a straight mathematics problem. Nothing more than that. We definitely do not want to lose the forest from the trees.*0053

*Okay. Let us get started. Our first example is going to be, so we will say let f(x,y) = x + y. Nice, simple function of 2 variables.*0062

*So, we want to find the max and min values of f on the square with coordinates + or -1, + or -1.*0080

*Okay, this is what this means. So, we have a square at... sorry... (1,1), (1,-1), (-1,1), (-1,-1).*0118

*So we have this square, and this is closed. This is definitely a closed and bounded set, so in the entire x, y plane, our domain is just what is inside here and what is on the boundary.*0135

*The theorem says that the function on a closed and bounded set contains a maximum and a minimum. Our job is to find that maximum and minimum, the points on there and the actual value of the function.*0151

*So, we have an interior and we have the boundary, so let us just go ahead and do what we do.*0165

*The first thing we do is, of course, take the partial derivatives, set them equal to 0. If they are equal to 0, then start with the local max's. In other words, the interior points.*0170

*So when I do df/dx, I get 1, and if I do df/dy, I also get 1, so as it turns out, this particular function on the entire x, y plane, there is nowhere... there is no critical point, because there is nowhere that the derivative is equal to 0.*0179

*What this tells me is that if there is a maximum and a minimum, which there is, it is guaranteed by the last theorem, by the last lecture, that maximum and min occur somewhere on the boundary.*0203

*It does not occur anywhere inside. It is going to happen somewhere on here.*0214

*Now, I am going to end up having to evaluate the function on the boundary. That is what is going on here.*0221

*Well, let us just start. So what I am going to do is I am going to call this line segment on the left, I am going to call it C1, and I am going to call this line segment here C2, and I am going to call this one C3, and I am going to call this one C4.*0229

*I am going to evaluate the values of the function all along all of these boundaries and see which number gives me the biggest value. That is how I do it.*0245

*So, let us go ahead and let us see. Let us check the boundary points.*0257

*Well, let us do C1 first. So C1, every single point on here has the coordinate (-1,y), okay?*0271

*So, f(-1,y) has a max at (-1,1), and the value of f(-1,1) is equal to 0.*0285

*Well, it has a min at (-1,-1) and the value at (-1,-1) equals -2. That is what I am doing.*0308

*I am just evaluating, now, everything on the boundary. Once I collect all of my values of all the different values of the function on the boundaries, the max's and the min's, I am just going to pick the biggest, and I am going to pick the smallest.*0325

*Now, let us try C2. So on here, the max was up here, and the min was here. Everything else was just in between. Again, we do not care about what was in between, we just about the max and the min, somewhere on there.*0340

*So, here, all of the points along this particular line have the y coordinate equal to -1, so f(x,-1). Okay.*0355

*So, f of all these points that have the same y coordinate, it is the x value that varies.*0372

*So, f(x,-1), it has a max at 1... no, wait... 1, 1... wait a minute... where I am I? let us see... 2... 1, 1, -1, oh, you know what? I actually... oops I think I made a little mistake here. Let me go ahead and reparameterize.*0377

*Let me go ahead and put C2 here, C3 here, and C4 here. So C2, I am actually going to be doing this line.*0411

*So, same thing. Instead of x,-1, it is actually going to be (x,1). So y value is 1 and x varies as we move from left to right.*0421

*So the maximum is going to be at (1,1), in other words, that point, and the value at (1,1) is going to be 2 and it is going to take its minimum value on this line over there.*0433

*So, it has a min at (-1,1), and f(-1,1), oh well when I put it in f, I get 0.*0450

*Let me go ahead and move on to the next page, and let me rewrite f again, so that I have it. It is going to be x + y, and I am also going to redraw my square real quickly, just because I would like to know what it is that I am looking at.*0468

*So I think we had C1, C2, C3, and C4. Okay. So now let us do C3. Now we are going to check the values along C3.*0489

*The coordinate of C3, the x value is going to be fixed at 1, and it is the y value that is going to vary.*0500

*So here, it is going to be (1,y), something like that. So f(1,y), well it has a max at (1,1) and f(1,1) is 2.*0509

*It has a min at (1,-1), which is this point and the value is 0. I am just doing this systematically. Just go through every single... cover the entire boundary, just make sure that all of the points are covered.*0530

*Now we will do C4, which is write down there, and the coordinates for C4, well the y value is fixed at -1, and it looks like the x value that is actually going to vary.*0553

*f(x,-1), that has a max at... it looks like that is the point at (1,-1), and the value of (1,-1) is equal to 0.*0565

*Then, it has a min at (-1,-1), which is that point. The value of (-1,-1) is equal to 2, or -2... sorry about that.*0593

*So, on the closed and bounded region, let us just call this s on that square, so on the closed and bounded region s, the maximum of f if I look through all of these values, the maximum is going to be 2, and the min is going to be -2. That is it.*0614

*Okay. Let us do another example here. Let us actually go do a different page. Example 2.*0644

*This time we want to find the extrema for f(x,y) = x ^{2} + 2y^{2} - x.*0656

*Let us see what is going to happen here. Well, let us go ahead and start by doing what we always do. Now notice that in this particular case they just said find the extrema of the function f(x,y) = x ^{2} + 2y^{2} - x.*0680

*They did not specify a domain, they did not actually specify some region. They did not say whether it was open or closed. When they do not do that, they are just saying over the entire x, y plane.*0694

*That is it. When they do not specify a region, your assumption is just that it is over the entire plane.*0705

*So, in this particular, we are going to be looking for absolute max, but it is actually going to end up being a local max, or an absolute min, or a local min, because we do not have a boundary that is keeping us contained.*0710

*Any max or min that takes place is going to end up being some critical point. That is what is going on here.*0726

*So, let us do what we always do. Let us go ahead and find the... so we will do df/dx, so the partial derivative with respect to x, it looks like it is going to be 2x-1.*0732

*The partial derivative with respect to y... it looks like it is going to be 4y.*0744

*So, we set each of those equal to 0, so we end up with y = 0 as the y point, and 2x-1, x is going to equal 1/2. There you go.*0753

*Our critical value... so let us write, our critical value equals the point (1/2,0). That (1/2,0), the derivative = 0. It is a critical point.*0765

*Now we just need to find out if it is a max, or if it is a min. So, here is where you are have to sort of stop and make some decisions about how you are going to proceed.*0779

*Are you going to do the same thing that you did before? Where you take the point (1/2,0), also it is going to be some like that and we can do what we did in one of the examples from the previous lesson.*0792

*We can hold 1 variable fixed and move a little bit to the left, move a little bit to the right and hold y fixed, or we can hold x fixed and we will move a little bit up and a little bit down, and we will check the value of the function around that critical point to see how the function behaves.*0802

*To see if all of the values around it are bigger than what the value of the function is there, or less than what the value of the function is there. To see if it is actually a maximum or minimum.*0821

*First of all, let us go ahead and find out what the value of the function is at (1/2,0).*0833

*So f(1/2,0) is going to equal... 1/2 ^{2} is 1/4, let us see... 0, this is 0... -1/2 equals -1/4, so the value of the function at the critical point is -1/4.*0839

*Here is where you stop and think about what you are going to do. Are you going to go through this process? Which you can, it is not a problem, you will get the answer. It is perfectly acceptable to do that.*0859

*But in this case, let us look at the function. We know the value is -1/4.*0870

*Well, take a look at the first two terms of this function. You have got x ^{2} and you have got y^{2}.*0876

*So, in this particular case, both of these terms, no matter what value of x you put in, no matter what value of y you put in, you are always going to end up with positive terms here.*0884

*Because they are positive terms, you are just subtracting this final x, you are always going to end up getting some number. In other words f(x) is always going to be bigger than f(1/2,0).*0898

*So these are positive. That is what is going on here. Just sort of by looking at this function and stopping and thinking about it, here is what you can do.*0911

*So, always positive, so... because this and this are squared terms... let me write it again actually.*0921

*So, f(x,y) = x ^{2} + 2y^{2} - x. This is always going to be a positive term. This is always positive. Therefore, f(x,y) is always going to be > or = to 1/4, which is the value of the function there.*0931

*So no matter if I move this way, this way, this way, this way, no matter which way I move, because these are both squared terms and they are both positive and it is always going to be a value that is bigger than -1/4.*0963

*Therefore, this is actually a minimum. In other words, all of the values around the function are going to end up being bigger... so it is going to be something like this. It is going to go that way.*0975

*If I hold the other variable fixed, it is going to go that way. So this is a minimum.*0988

*So, min at (1/2,0) = -1/4, so this is where the minimum takes place. The value of the minimum is -1/4, so that is it.*0995

*Again, you are welcome to go ahead and go through the process that we did of holding a variable fixed, moving a little this way, moving a little this way, you are going to get the same answer.*1013

*In this case it was nice where we had a particular function, look at it, stop and think for a second, and say, Okay, well both of those terms are positive therefore at that value it is always going to be -- the value of the function is always going to be -- bigger than -1/4.*1021

*That is what is happening. Let us do another example. Let us see, example 3.*1038

*Find the max and min of f(x,y) = (x ^{2} + y^{2})^{-1}.*1055

*Okay, so slightly more complicated function here. In the region -- we are going to add one more bit of complication here -- in the region x-2 ^{2} + y^{2} is less than or equal to 1.*1075

*Now I have not only specified a function, I have specified a region except now the region is given as an equation, so this is an equation of a circle that is centered at (2,0).*1096

*Let us go ahead and draw this out because we can. Any time you can actually draw a picture is is a good idea. So this is 1, this is 2, so this is the center of the circle. There is that, there is that, and there is 3.*1107

*So, we have a circle and it is less than or equal to 1 of radius 1, okay? It is closed, this is a closed and bounded region.*1122

*What we are saying here is on this region, all the interior points including the boundary points, what is the maximum value that this function obtains, and what is the minimum value that this function obtains. That is all that we are doing here.*1133

*Well, let us just start and see where we go. So, let us just take the df/dx.*1148

*When we take the first partial derivative, this is going to equal -x ^{2} + y^{2} × the derivative of what is inside, which is 2x.*1154

*df/dy = -x ^{2} + y^{2} × 2y. Again, we are just taking the derivative of this function, this is a composite function, so we take -1 × this... -2 power, sorry about that, and then of course we take the derivative of what is inside.*1171

*Since we are differentiating with respect to y, it is just 2y here. The first one we differentiated with respect to x.*1194

*Not notice, take a look at this. Well, for one thing, the function is not even defined at (0,0). You cannot even have that because this is... okay, let me rewrite this, so this is going to be (-2x/x ^{2} + y^{2})^{2},*1202

*This is going to be -2y/(x ^{2} + y^{2})^{2}.*1220

*First of all, this function is not even defined at (0,0), however for every other value of x and y, the derivative never equals 0.*1229

*So, df/dx does not equal 0 anywhere on the domain, anywhere in the region.*1240

*Not that we were concerned with (0,0) because (0,0) is not even part of this. But anywhere on this region, the derivative does not equal 0, and neither does df/dy.*1250

*df/dy does not equal 0 anywhere on the region in question, because the derivative does not equal 0 anywhere on that region, we know that it is not going to take place anywhere on the interior.*1260

*So, the max and min of this function is going to take place on the boundary. It is the boundary that we need to concern ourselves with.*1276

*Here is where it starts to get kind of interesting. So, let us go through this very, very carefully.*1282

*Now, let me go ahead and move on to another page. I am actually going to redraw it so that I have it available for me.*1290

*I have this 1, 2, 3, just do a quick drawing... so, I have that. The unit circle centered at 2.*1298

*So, the boundary is x - 2 ^{2} + y^{2} = 1. That is the boundary. The region was < or = to 1, but the boundary is at 1.*1308

*How can we find the values of the function -- let me rewrite the function -- so, find values of f(x,y), which is equal to x ^{2} + y^{2}... let me write it actually as a fraction so that we have positive exponents.*1328

*1/(x ^{2} + y^{2})^{1}. So that is actually what we are looking... how can we find the values of this function on this circle?*1356

*The value of this function on this circle. This is a composite function. I need to be able to take this, I need to be able to form f, so this is some... I need to be able to form f(g). This is g, this is f. I need to form the composite function.*1370

*Here is what I am going to do. I am going to go ahead and parameterize this circle, and express it as a function of one variable, t.*1390

*Let us go ahead and parameterize this circle. Well, I know what the parameterization of a circle is in general.*1397

*So, a general circle, the parameterization is just c(t) = cos(t)sin(t), right? That is the parameterization of a unit circle centered at the origin. But this one is not centered at the origin, it is centered at 2.*1403

*My parameterization for this function here, x - 2 ^{2} + y^{2} = 1, looks like this. Equals 2 + cos(t)sin(t).*1420

*If you were to put 0, pi/2, 3pi/2, and 2pi into t, you will get this point, this point, this point, and this point. That is your parameterization moving in that direction.*1434

*This is a parameterization of this function. All I have done is taken an implicit function, and I have expressed it parametrically.*1448

*Now I can actually form f(c(t)). That is what is actually going on here. So, c(t) is this, this is f, I am going to form f(c(t)).*1458

*f(c(t)) is equal to, well, it is going to be 1/x ^{2}, well x is this. That is x, that is y, so it is going to be 2 + cos(t)^{2} + sin(t)^{2}.*1470

*Now we are going to go ahead and actually multiply everything out. It is going to equal 1 over, well, 2 × 2 is 4, 2cos(t)2cos(t) is + 4cos(t), + cos ^{2}(t) + sin^{2}(t). That is the denominator.*1500

*Well, cos ^{2}(t) + sin^{2}(t) is the pythagorean identity. That is equal to 1, so 4 + 1.*1523

*So my f(c(t)) = 1/5 + 4cos(t).*1530

*There we go. I found a way of expressing my function as a single variable, a function of t, this is my function. This is f(x,y).*1546

*Now, all I need to do is find the values of t that make this a maximum, and find the values of t that make this a minimum.*1557

*Well, this is actually really, really simple here. Notice this cos(t).*1565

*Let us just... well, let us just stop here. Let us just say f is a max, the function chooses its maximum value when the denominator is the smallest and f achieves this number here.*1572

*Achieves a minimum value, a smallest value, when the denominator is the biggest, right?*1600

*This is the function, big denominator, small value. Small denominator, big value. That is what is going on here.*1615

*Now let us go ahead and find what this is. Well, what do we know about cos(t)?*1625

*Well, cos(t) has its biggest value of 1, and its smallest value at -1. It ranges between -1 and 1.*1629

*Therefore, it is going to achieve... so cos(t) = 1, when t=0... so we want this value here. We want to achieve the biggest and smallest value. Well, we want to achieve 1, and we want to achieve -1.*1640

*Okay. It is equal to 1 when we have cos(t) = 0, therefore when t = 0, that means the cos(0) is 1. That means 1 + 5 × 4 × 1. It is going to be 1/9.*1666

*Therefore at t=0, that is when it is going to achieve its minimum value.*1693

*Now the minimum value of cos(t)... cos(t) achieves -1 when it is equal to pi. So when t = pi, the cos is equal to -1.*1699

*Well -1 × 4 is -4, 5 -4 is 1, so now it achieves its maximum value there. So let us actually just write this out.*1711

*When t = 0, c(t) = 2 + cos(0)sin(0) = (3,0). That is the point at which it is going to achieve one of its extrema value.*1728

*Well, f(c(t)) is going to equal 1/5 + 4, which equals 1/9. That is the minimum.*1752

*Now, when t = pi, cos(t) = -1. c(t) = 2 + cos(pi) × sin(pi), which is equal to (1,0).*1768

*When I put (1,0) into that function, f(c(t)) = 1/1, because remember, we had 1/5 + 4 × cos(t). 5 + 4 × cos(t).*1795

*When t = 0, it is 1/9, so it is 1/9. The value of the function at the point (3,0) is equal to 1/9.*1821

*When t = pi, that is when it is at its smallest. The smallest value of cos is when t = pi.*1833

*When t = pi, well cos(pi) is -1, 4 × -1 is -4, 5 - 4 is 1. So the value of the function is 1.*1840

*At the point (1,0), the value of the function is 1. That is the maximum.*1851

*At the point (3,0), the value of the function is 1/9. That is the minimum.*1859

*Here is what we have done. Here is our circle. 1, 2, 3, we have a circle here. That looks like an ellipse, so let us make it look a little bit like a circle.*1863

*We have this circle, so at the point (1,0), here is where it achieves its maximum value of the function. At the point (3,0), here is where the function actually achieves its minimum value.*1873

*We did this by just taking our function, parameterizing it, putting it into our region, parameterizing the curve, putting that parameterization, forming (f(c(t)) like we have been doing for a while, and then we just found the values of t that make the function a maximum or a minimum.*1886

*The only real problem here is keeping track of all of the mathematics. There is a lot on the page.*1909

*There are a lot of numbers floating around, there is a lot of points floating around. The only thing that you have to do is make sure that you differentiate between the point at which the maximum or minimum is achieved.*1914

*In other words, these points. Those are the points at which they are achieved, and the actual value of the function at those points, that is here, and that is here.*1925

*Again, we want to rely on the techniques that we have at our disposal, but really what we want to rely on is our intuition and what we know about functions and how they behave and how numbers work.*1935

*When you get to this point, you can reason out the rest. When this is the biggest, the number is going to be the smallest. When this is the smallest, this number is going ot be the biggest.*1947

*You just need to go step by step and make sure that everything is there. Okay. I hope that this has worked out and everything is well.*1958

*Thank you for joining us here at educator.com and multivariable calculus. We will see you next time.*1963

1 answer

Last reply by: Professor Hovasapian

Tue Jun 3, 2014 7:19 PM

Post by Juan Castro on May 22, 2014

Raffi,

This lecture could have been hard but you made it seemingly easy.

Great teaching style. Thanks!

1 answer

Last reply by: Professor Hovasapian

Fri Apr 12, 2013 12:32 AM

Post by Jawad Hassan on April 11, 2013

the last example was so big, do you think this is normal problems in a exam? or is too long for a exam problem?