If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises. 

A critical point of a function y = f(x) is a point at which f ' (x) = 0.  If at the critical point  f ''(x) < 0, the derivative is decreasing and the critical point is a maximum; if f ''(x) > 0 the derivative is increasing and the critical point is a maximum.  If f '' (x) = 0, then the test is inconclusive and additional analysis is necessary.

At a point (x0, y0) the function z = f(x, y) yields two functions, z = f(x, y0) and z = f(x0, y).  The first is a function of x and the second is a function of y. 

It is possible that both the x and y functions have a maximum at (x0, y0), in which case we would tend to expect that f(x, y) has a maximum at this point, or that both have minima at (x0, y0),  in which case we would tend to expect that f(x, y) has a minimum.  As we will see, though, it's not quite that simple.

It is also possible that one function has a maximum, while the other has a minimum.  In this case we expect that f(x, y) has a 'saddle point' at (x0, y0).

A saddle point can also occur even if the x and y functions match, in that both have maxima, or both have minima.  This is because the x and y second derivatives indicate the behavior only in the x and y directions.  They don't tell us, for example, what happens if we move in a direction at 45 degrees relative to the x and y directions.  It is entirely possible that, for example, our x and y functions are each have a minimum at a point, with the graph rising if we move from our point in either the x or the y direction, but that the graph falls as we move in a direction at 45 degrees.

The test for max or min of a function of two variables is as follows:

• f_x and f_y are the derivatives in the x and y directions.  If we do have a relative max or a relative min, both f_x and f_y must be zero.  We therefore find the critical points of f(x, y) by solving the equations f_x (x, y) = 0, and f_y (x, y) = 0.
• We then test the second derivatives f_xx and f_yy, which indicate the concavity of our graph in the x and y directions, at each of our critical points.  If the concavity at a critical point is the same in both directions, then it is possible but not certain that our function f(x, y) has a relative max or min at that critical point.
• Note that the concavity is the same in both x and y directions if f_xx and f_yy have the same sign (and neither is zero).  If we multiply two numbers of the same sign, we know that their product is positive.  So the requirement that both have the same sign, with neither being zero, is equivalent to saying that f_xx * f_yy > 0.
• To see whether we do have a max or a min, we also have to test the behavior of the function in the 45-degree direction.  We do this for a critical point by finding the 'mixed partial derivative' f_xy.
• If f_xx * f_yy - f_xy ^2 is still greater than zero, then we can be sure that the 45-degree direction doesn't mess thing up.  In this case we don't have a saddle point, and we have a relative max or relative min, depending on whether f_xx and f_yy are both negative (giving us a relative max) or both positive (giving us a relative min).

Question: 

`q001.  Let (x0, y0) = (2, 3), and let f(x, y) = 2 x^2 - 7 xy + 4 y^2 - 8.  The given point (x0, y0) is not a critical point for the function, as you will demonstrate:

• What is the function z = f(x, y0)?
• This function f(x, y0) represents the intersection of the graph of f(x, y) with the plane y = y0.  You should sketch its graph.  What is the derivative of this function with respect to x?  What are the critical points of this function?  Does the function f(x, y0) have a max, a min or an inflection point at your critical point(s)?

• What is the function z = f(x0, y)?
• This function f(x0, y) represents the intersection of the graph of f(x, y) with the plane x = x0.  You should sketch its graph.  What is the derivative of this function with respect to y?  What are the critical points of this function?  Does the function f(x0, y) have a max, a min or an inflection point at your critical point(s)?

• Do the critical points of f(x, y0) coincide with the critical points of f(x0, y)?  Do you therefore conclude that (x0, y0) is or is not a critical point of the function f(x, y)?

Your solution: 

Confidence rating:
 

The x derivative of this function is 4 x - 21.

The critical point occurs when the derivative is zero, which occurs at x = 5.25 (this had better be so because the vertex of a parabolic graph is a critical point, and occurs on the axis of symmetry)

The second derivative is 4, so the critical point yields a relative minimum.

A relative minimum in the x direction therefore occurs when (x, y) = (5.25, 3).

f(x0, y) = f(2, y) = -14 y + 4 y^2 - 4.

This is also a parabola, whose axis of symmetry is the line y = 7/4.

The y derivative of the function is -14 + 8 y.

The critical point occurs when the derivative is zero, which is easily found to occur when y = 7/4 (again this should, and does, agree with the fact that the axis of symmetry is y = 7/4).

Again the second derivative is positive, indicating that the critical point yields a relative minimum.

A relative minimum in the y direction therefore occurs when (x, y) = (2, 7/4).

The points (5.25, 3) and (2, 7/4) are not the same.

The relative minimum for the x = 2 'slice' of the graph of f(x, y) therefore does not occur at the same point as the relative minimum for the y = 3 'slice'.

If (x0, y0) was a critical point for f(x, y), the critical points for the x = x0 and y = y0 'slices' would coincide, and either both would be relative maxima or both would be relative minima.  (As we will see, even if this condition holds, a 'saddle point' is still possible).

`q002.  Find f_x and f_y for the function f(x, y) = 2 x^2 - 7 xy + 4 y^2 - 8.  Set both of your expressions equal to zero.  This will give you two simultaneous equations in x and y. 

• Solve the two equations to find the critical point(s).

Your solution: 

Confidence rating:
 

Find f_xx, f_yy and f_xy.

At each of your critical points, determine whether f_xx is positive or negative (or zero), and determine the same for f_yy.  Let each critical point in turn be (x0, y0).

• Do your results indicate that the x function f(x, y0) has a max or a min?  If so, which is it?
• Do your results indicate that the y function f(x0, y) has a max or a min?  If so, which is it?

At each of your critical points find the value of f_xy, the value of f_xx * f_yy and the value of f_xx * f_yy - f_xy ^2.

• Using your values conclude whether the function has a max, a min or a saddle point at the critical point.

f_y = 8 y - 7 x.

The two are both equal to zero when

4 x - 7 y = 0 and

8 y - 7 x = 0.

Solving simultaneously we get x = 0 and y = 0.  Thus our critical point is (x0, y0) = (0, 0).

f_xx = 4 and f_yy = 8.  Both are positive.  So both functions f(x0, y) = f(0, y) = 4 y^2 and f(x, y0) = f(x, 0) = 2 x^2 have minima at (0, 0).  A relative minimum is therefore possible at (0, 0).

However we also have to check out f_xy.  In this case we find the f_xy = -7.

Testing the expression f_xx * f_yy - f_xy ^2 we find that its value is 4 * 8 - 7^2 = 32 - 49 = -17.  It follows that our critical point (x0, y0) = (0, 0) is a saddle point for this function. 

The intersection of the graph with the y = 0 plane yields a curve which has a minimum, as does the intersection with the x = 0 plane.  However in the plane x = y, for example, it gives a maximum.