f(x, y) = 1 sin(2x) sin(2y) and determine whether each is a maximum, minimum, or saddle. Solution: Critical points occur when f(x, y) = 0.

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1 Some Answers to Exam Name: (4) Find all critical points in [, π) [, π) of the function: f(x, y) = sin(x) sin(y) 4 and determine whether each is a maximum, minimum, or saddle. Solution: Critical points occur when f(x, y) =. We have: f x (x, y) = cos(x) sin(y) f y (x, y) = sin(x) cos(y) The first equation tells that x = π, 3π or that y =, π. That is either we have x = π, 3π or y =, π. Notice that all 4 4 these points are in the domain of x and y. The second equation tells us that y = π, 3π or that x =, π. That is either we have y = π, 3π or x =, π. Notice that all 4 4 these points are in the domain of x and y. To find critical points we get to pick one of x or y from solutions to the first equation and the other one from solutions to the second equation. Therefore we get the following critical points (x, y): A. (, ) B. (, π/) C. (π/, ) D. (π/, π/) E. (π/4, π/4) F. (π/4, 3π/4) G. (3π/4, π/4) H. (3π/4, 3π/4) To determine if these are maximima, minima, or saddles we need to look at the Hessian, the matrix of nd partials/

2 f xx (x, y) = sin(x) sin(y) f xy (x, y) = cos(x) cos(y) f yx (x, y) = cos(x) cos(y) f yy (x, y) = sin(x) sin(y) So at each critical point we have the following Hessians: A. B. C. D. E. F. G. H. ( ) ( ) Finally, calculate the determinate and trace of each Hessian. If the determinate is negative we have a saddle. If it is positive, the trace being positive tells us we have a minimum and the trace being negative tells us we have a maximum.

3 3 Crit Pt Det Trace Type A - Saddle B - Saddle C - Saddle D - Saddle E - Max F Min G Min H - Max (9) Let R be the object defined by the following bounds: y y x + y ln(y + y) z ln(y + y) + 3y Use a triple integral to comput the volume of R. Hint: Consider the change of variables: x = u + v y = v z = ln(v + v) + vw Solution: We will use the change of variables formula. To do that we need to compute the bounds of the corresponding region in terms of u, v, w and we need to compute the determinate of the jacobian of the transformation. The change of coordinates formula is: f(x, y, z)dxdydz = f(u, v, w) detdt dudvdw R R First compute the new region R : v v u + v + v ln(v + v) ln(v + v) + vw ln(v + v) + 3v The idea is now to do some simple algebra to rewrite the bounds so that u, v, w are in the middle of their respective inequalities. **There is one issue here that needs to be addressed (though it didn t affect the grading). We ll discuss it at the end of the solution.

4 After some algebra we get: v u w 3 Now we can find detdt. To do that write down the matrix of partial derivatives from the equations for the transformation: DT = u(u + v ) / v(u + v ) / v+ + w v v +v Thus detdt = uv u. Now we can write down the integral: +v uv u + v dudvdw. The inside integral can solved by substitution r = u. We ll use this substitution again later: uv u dudvdw = +v v v ) / drdvdw = + v v dvdw = + sds 3 v3 dw = 3 ( + s)3/ 3 dw = 3 ()3/ dw = 3 8 **Now for the tricky bit. Notice that in the original bounds for R, y was allowed to equal. In the bounds for z, however, we have ln(y +y) which is undefined when y =. Here s a way around that: Let R c for > c > be the object with bounds: c y y x + y ln(y + y) z ln(y + y) + 3y 4

5 The volume of R is then equal to lim c + R c dxdydz. Notice that this solves the division by problem in calculating the limits for u. Now go through the above work. Your final answer will have c in it. Now take the limit as c +. You ll get the answer above. 5