Calculus Practice Exams

Time: 1 hour


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

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Answer. The critical points are

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Since

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(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

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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

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that is the closest to the origin.

Answer. Set

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Then by the method of Lagrange multipliers if (x,y,z) is the point which is the closest to the origin, we must have

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where tex2html_wrap_inline137 and tex2html_wrap_inline139 are real numbers. We get

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Playing around with the system knowing that x-y+z=4 and x+y-z=8, we get

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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

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Another way to prove this is to parametrize the line

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which gives

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Plug these into f(x,y,z) to find

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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

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under the constraints

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Answer. Set

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Then by the method of Lagrange multipliers if (x,y,z) is a local maxima (or minima), we must have

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where tex2html_wrap_inline137 and tex2html_wrap_inline139 are real numbers. We get

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Playing around with the system knowing that x-y=0 and z=y-2, we get

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Since

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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

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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

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over the plane region G bounded by the lines y = x, y = -x and x = 4

Answer. Clearly we have

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(see the picture below of the set G)

Therefore, we have

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Easy calculations give

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Hence

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Problem 5. Evaluate

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Answer. Set

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(see the picture below of the set G)

In polar coordinates, we get

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Therefore we have

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Since

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we get

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