Download Fall 2009 MA 227-6B Test 3: Finding Extrema, Gradient, Lagrange Multipliers, and Integrals and more Exams Advanced Calculus in PDF only on Docsity!
FALL 2009 — MA 227-6B — TEST 3
NOVEMBER 18, 2009
Name:
There are 7 problems on this test. The number of points to earn is indicated for each problem. Partial credit is awarded where appropriate. Your solution must include enough detail to justify any conclusions you reach in answering the question. Points will be awarded for the correct reasoning.
(1) (21 points) Find the local maxima, minima, and saddle points of the function g(x, y) = x^2 + 2y^2 + x^2 y. Solution: Local extrema and saddle points occur at critical points, i.e., points where the gra- dient is zero. The gradient of g is ∇g = 〈 2 x(1 + y), 4 y + x^2 〉. The first component is zero either if x = 0 or else if y = −1. In the first case, when x = 0, the second component is 4y which is zero when y = 0. Hence (0, 0) is a critical point. In the second case, when y = −1, the second component is x^2 − 4 which is zero when x = ±2. Hence (2, −1) and (− 2 , −1) are critical points. The nature of a critical point is (often) decided by the second derivative test. We get gxx = 2 + 2y, gxy = gyx = 2x, and gyy = 4. Hence D(x, y) = gxxgyy − g^2 xy = 8(1 + y) − 4 x^2. Since D(0, 0) = 8 > 0 the point (0, 0) is an extremum. Since gxx(0, 0) = 2 > 0 (note that also gyy = 4 > 0) the point (0, 0) is a minimum. Since D(± 2 , 1) = − 16 < 0 the points (2, −1) and (− 2 , −1) are saddle points.
1
(2) (21 points) Use Lagrange multipliers to find maximum and minimum values of the function f (x, y) = 2x + y + 5 on the circle x^2 + y^2 = 1. Where do they occur? Solution: Only points on the circle are to be considered. This constraint is expressed as g(x, y) = 0 where g(x, y) = x^2 + y^2 − 1. The method of Lagrange multipliers re- quires to find the solutions of the following system of algebraic equations: ∇f = λ∇g, g = 0. In this case ∇f = 〈 2 , 1 〉 and ∇g = 〈 2 x, 2 y〉 so that the system is 2 = 2λx 1 = 2λy 1 = x^2 + y^2. From the first equation (for instance) we get that λ can not be zero (otherwise the first equation would be 2 = 0 which is absurd). Hence x = 2/(2λ) and y = 1/(2λ). Therefore x = 2y. Using this in the third equation gives 1 = (2y)^2 + y^2 = 5y^2. There are two solutions: y 1 = 1/
5 and y 2 = − 1 /
- Recalling x = 2y gives two points to consider: P 1 = (2, 1)/
5 and P 2 = −(2, 1)/
- The value of f at P 1 is 5 /
5 + 5 and the value of f at P 2 is − 5 /
5 + 5. The former is the larger of the two. Hence the maximum occurs at P 1 and the minimum at P 2.
(4) (13 points) Use polar coordinates to evaluate the integral
D xy
(^2) dA where D is the left half of the disk of radius 3 centered at the origin. Solution: For polar coordinates we have x = r cos(θ) and y = r sin(θ). The points in the left half plane have values of θ between π/2 and 3π/2. The points in the disk of radius 3 centered at the origin have values of r between 0 and 3. We must not forget that dA = rdrdθ. Hence ∫ ∫
D
xy^2 dA =
∫ (^3) π/ 2
π/ 2
0
r cos(θ)(r sin(θ))^2 rdrdθ
∫ (^3) π/ 2
π/ 2
cos(θ)(sin(θ))^2
0
r^4 drdθ =
r^5 5
∣r= r=
∫ (^3) π/ 2
π/ 2
cos(θ)(sin(θ))^2 dθ.
To evaluate the θ-integral we substitute u = sin(θ), du = cos(θ)dθ. sin(π/2) = 1 will be the lower limit of the integral while sin(3π/2) = −1 will be the upper limit. Hence ∫ (^3) π/ 2
π/ 2
cos(θ)(sin(θ))^2 dθ =
1
u^2 du =
u^3 3
u=− 1 u=
Together with the previous result we get ∫ ∫
D
xy^2 dA = −
(5) (13 points) Find the absolute maximum and minimum as well as the points where they occur for the function f (x, y) = 3 + xy over the disk of radius
8 centered at the origin. Solution: We have to investigate both the interior and the boundary of the disk. For the interior we find the critical points of f , i.e., the zeros of ∇f. Since ∇f = 〈y, x〉 there is precisely one critical point, the origin. Note that f (0, 0) = 3. The boundary of the disk is described by the constraint equation g(x, y) = 0 where g(x, y) = x^2 + y^2 − 8. Here we use Lagrange multipliers. This method requires to find the solutions of the following system of algebraic equations: ∇f = λ∇g, g = 0. In this case ∇f = 〈y, x〉 and ∇g = 〈 2 x, 2 y〉 so that the system is y = 2λx x = 2λy 8 = x^2 + y^2. From the first two equations we get x = 4λ^2 x. If x = 0 then y = 0 and 0 = 8 so that x cannot be zero. Therefore we get 4λ^2 = 1 and 2λ is either 1 or −1. First let 2λ = 1, so that x = y. Then the third equation is 2x^2 = 8 which holds when x = 2 or x = −2. We have two points to consider: (2, 2) and (− 2 , −2). At both points f assumes the value 7. Now let 2λ = −1, so that x = −y. Then the third equation is again 2x^2 = 8. Now the two points are (2, −2) and (− 2 , 2). At both of these f assumes the value −1. Comparing the three values −1, 3, and 7 we see that the minimum is −1 attained at (2, −2) and (− 2 , 2) and the maximum is 7 attained at (2, 2) and (− 2 , −2).
(7) (4 points) The ellipse E defined 4x^2 + 9y^2 ≤ 36 can be transformed into a circle by a change of variables. Perform such a change of variables to find the area of the ellipse. Solution: Dividing by 36 we may express ellipse as (x
3
(y
2
Thus we substitute u = x/3 and v = y/2, i.e., x = 3u and y = 2v to get the disk u^2 + v^2 ≤ 1. This transformation has Jacobian
J =
Then
Area(E) =
E
1 dA(x, y) =
D
JdA(u, v) = J
D
1 dA(u, v) = JArea(D)
where D is the disk u^2 + v^2 ≤ 1 which has area π. Hence the area of E is 6π.
