## Calculus Practice Exams

Time: 1 hour

Problem 1. Find all local maxima and minima of the function Answer. The critical points are Since (1,2,3) and (-1,2,3) are possible candidates for local minima and (0,2,3) is a possible candidate for local maxima. Note that by completing the squares, we get which implies that (1,2,3) and (-1,2,3) are not only local minima but global minima. The point (0,2,3) is not a local maxima.

Problem 2. Use the method of Lagrange multipliers to find the point on the line of intersection of the planes that is the closest to the origin. Then by the method of Lagrange multipliers if (x,y,z) is the point which is the closest to the origin, we must have where and are real numbers. We get Playing around with the system knowing that x-y+z=4 and x+y-z=8, we get This leads us to conclude that a good candidate for the minimum of f(x,y,z) is (6,1,-1) with f(6,1,-1)=38. The shortest distance to the origin is therefore Another way to prove this is to parametrize the line which gives Plug these into f(x,y,z) to find It is now clear that the minimum is achieved at t=-1 and the minimum value is 38...

Problem 3. Use the method of Lagrange multipliers to find local maxima of under the constraints  Then by the method of Lagrange multipliers if (x,y,z) is a local maxima (or minima), we must have where and are real numbers. We get Playing around with the system knowing that x-y=0 and z=y-2, we get Since then f(x,y,z) has a local maxima at (0,0,-2). Another way to see this set x=y and z=y-2 in the definition of f. We will get It is very easy to see that y=0 is a local maxima and y=4/3 is a local minima....

Problem 4. Find the integral of over the plane region G bounded by the lines y = x, y = -x and x = 4

Answer. Clearly we have (see the picture below of the set G) Therefore, we have Easy calculations give Hence Problem 5. Evaluate  (see the picture below of the set G) In polar coordinates, we get Therefore we have Since we get [Calculus] [CyberExam] S.O.S MATH: Home Page

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