## Calculus Practice Exams

Time: 1 hour

Problem 1. Find all local maxima and minima of the function

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

(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

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