Maxima and Minima

1. Find all critical points of f(x, y) = x2 + y2.

Solution. We want fx = fy = 0. We calculate fx = 2x, fy = 2y , and the only way that this can be

true is for x and y to both be 0. Thus, the only critical point is (x, y) = (0, 0) .

2. Find all critical points of f(x, y) = 5− x3 + y3 + 3xy.

Solution. We want fx = fy = 0, and we can calculate fx = −3x2 + 3y, fy = 3y2 + 3x. So, we want−3x2 + 3y = 0 and 3y2 + 3x = 0. The first equation tells us that x2 = y; plugging this into the second,we get that x4 + x = 0 = x(x3 + 1), so x = 0 or −1 . Plugging this back into the first equation gives

y = 0 or 1, so (0, 0) and (−1, 1) are the only two critical point.

3. Find all critical points of f(x, y) = xy2 − x2 − 2y2 and determine whether each is a local minimum,local maximum, or saddle point.

Solution. We want ∇f = ~0, and we can calculate ∇f = 〈y2 − 2x, 2xy− 4y〉. So, we want y2 − 2x = 0

and 2xy− 4y = 0. The first equation tells us that x = y2

2 ; plugging this into the second equation gives

y3 − 4y = 0. We can factor this as y(y − 2)(y + 2) = 0, so y = 0, 2, or −2. Since x = y2

2 , this gives usthree points, (0, 0) and (2,±2).

To classify our three critical points, we need to calculate the discriminant fxxfyy − f2xy at each point.

Let’s first just calculate it in terms of x and y: fxx = −2, fxy = 2y, and fyy = 2x−4, so fxxfyy−f2xy =

−4(x− 2)− 4y2. So:

• At (0, 0), the discriminant is 8, which tells us that (0, 0) is either a local minimum or a lo-cal maximum. To decide which, we can use the fact that fxx = −2 < 0, which tells us that

(0, 0) is a local maximum .

• At (2, 2), the discriminant is −16, so (2, 2) is a saddle point .

• At (2,−2), the discriminant is −16, so (2,−2) is a saddle point .

4. Find the points on the cone z2 = x2 + y2 that are closest to the point (4, 2, 0).

Solution. The distance from a point (x, y, z) to (4, 2, 0) is√

(x− 4)2 + (y − 2)2 + z2. Finding minimalvalues of distance is the same as finding minimal values of the square of the distance. So we letf = (x− 4)2 + (y − 2)2 + z2 and search for the minimum of f .

Before finding the critical points, notice that if (x, y, z) is on the cone, z is dependent on x, y, so wesubstitute z2 = x2 + y2 in f then f = f(x, y) = (x− 4)2 + (y − 2)2 + x2 + y2 is a function of two freevariables. Next, fx = 2(x − 4) + 2x = 4x − 8, fy = 2(y − 2) + 2y = 4y − 4. Therefore fx = fy = 0implies x = 2, y = 1. D = fxxfyy − f2

xy = 4 · 4 − 02 = 16. The point x = 2, y = 1 is a minimum.

z2 = 22 + 12 = 5, z =√

5 or −√

5.

Conclusion: (2, 1,√

5) and (2, 1,−√

5) are the closest point to (4, 2, 0) on the cone.

5. Find the absolute maximum and minimum values of f(x, y) = y2 − x2 on the square |x| ≤ 1, |y| ≤ 1.

Solution. Since |x| ≤ 1, |y| ≤ 1 is a closed bounded region in R2, we know that f(x, y) must attainits maximum and minimum values on this region. Remember that the basic strategy is to check the

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critical points and boundary separately. That is, we’ll find all the critical points, and then we’ll find allthe points on the boundary where the absolute minimum or maximum might occur; after we’ve donethat, we’ll plug each point into f to see which gives the highest value and which gives the lowest.

The critical points are where ∇ = ~0. Since ∇f = 〈−2x, 2y〉, the only critical point is (0, 0).(1)

Now, we look at the boundary. The boundary is composed of four separate pieces, and we’ll look ateach one individually:

• Let’s look at the piece where x = 1. Here, f(x, y) = f(1, y) = y2 − 1, and we are concerned withy in the interval [−1, 1]. So, we want to maximize and minimize y2 − 1 on the interval [−1, 1].You can do this using calculus or just by looking at the graph of y2 − 1; either way, you shouldfind that it’s biggest when y = ±1 and smallest when y = 0. This gives us three candidate points,(1, 1), (1,−1), and (1, 0).

• On the piece where x = −1, f(x, y) = f(−1, y) = y2 − 1, and we again want to maximize andminimize y2− 1 on the interval [−1, 1]. The maximum is when y = ±1 and the minimum is wheny = 0. This gives us three more candidate points, (−1, 1), (−1,−1), and (−1, 0).

• On the piece y = 1, f(x, y) = f(x, 1) = 1 − x2, and we want to maximize and minimize this onthe interval [−1, 1]. The maximum occurs when x = 0, and the minimum occurs when x = ±1,so we get three more candidates, (0, 1), (−1, 1), and (1, 1).

• On the piece y = −1, f(x, y) = f(x,−1) = 1 − x2, and we get three more candidates, (0,−1),(−1,−1), and (1,−1).

So, we now have a big list of points where the absolute minimum and maximum might occur: (0, 0),(1, 1), (1,−1), (1, 0), (−1, 1), (−1,−1), (−1, 0), (0, 1), (−1, 1), (1, 1), (0,−1), (−1,−1), and (1,−1).We evaluate f at each of these points, and we find that the absolute maximum occurs at (0,±1), wherethe function is equal to 1. The absolute minimum occurs at (±1, 0), where the function is equal to−1.

6. Find the absolute maximum and minimum values of 2x3+y4 on the unit disk D = {(x, y) |x2+y2 ≤ 1}.Solution. Let f(x, y) = 2x3 + y4, Since D is a closed and bounded region and f is continuous, weknow f must have attain global maximum and minimum in D.

Step 1: Find critical points: fx = 6x2, fy = 4y3, and hence the only critical point is (0, 0) and it’sindeed in D. f(0, 0) = 0.

Step 2: On the boundary x2 + y2 = 1,

f(x, y) = 2x3 + y4 = 2x3 + (1− x2)2 = x4 + 2x3 − 2x2 + 1.

Thus we regard it as a function of only one variable x and find extremum. This can be done by takingderivative f ′ = 4x3 + 6x2 = 4x and set it equal to zero. f ′ = 2x(2x− 1)(x + 2) therefore x = 0 or 1/2

(x = −2 is not in D). The correspoding points and values are f(0,±1) = 1, f( 12 ,±

√32 ) = 13

16 . Don’tforget about the endpoint x = −1 and x = 1, they are f(−1, 0) = −2, f(1, 0) = 2.

Step 3: After comparing all candidates, we find f(−1, 0) = −2 is the global minimum and f(1, 0) = 2is the global maximum.

(1)You can use the Second Derivative Test to see what type of critical point (0, 0) is, but it’s not necessary to do so since weare really just compiling a list of candidate points that we will check later.

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