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EXAMPLES FOR SECTION 13.8
EXAMPLE--Tangent Planes and Max-Min Find the points on the graph of the function below where the tangent plane is horizontal. f(x,y) = x3 - 3xy + y2 DPGraphPicture (z-axis scale: 1 unit = 10) DPGraphPicture with the 2 horizontal tangent planes Since the surface can be represented by F(x,y,z) = f(x,y) - z = x3 - 3xy + y2 - z = 0 and grad F will give a vector normal to the surface we need to find points where grad F = < 0 , 0 , c > where c is any nonzero constant. grad F = < fx , fy , -1 > so we are looking for points on the surface where fx = fy = 0. fx = 3x2 - 3y = 0 fy = -3x + 2y = 0
The solutions to this system are (0,0), and (3/2,9/4). f(0,0) = 0 and f(3/2,9/4) = -27/16 so the points on the surface where the tangent plane is horizontal are (0,0,0) and (3/2,9/4,-27/16). The picture on the left below shows the graph of the surface around the point (0,0,0) and the picture on the right shows the graph around the point (3/2,9/4,-27/16). Click on each graph to see an animation that will include the tangent plane at the critical point on that graph. To determine the nature of these points where the tangent plane is horizontal we can apply the Second Partials Test.
Now let's consider the problem of finding the absolute minimum and absolute maximum value of the function over the square region of the xy-coordinate plane described by the x-interval [-1,3] and y-interval [-1,3]. See the picture on the right above or this DPGraph Picture (z-axis scale: 1 unit = 10). The two critical points found above yield -27/16 as a possible minimum and 0 as a possible maximum. We must now evaluate the function on the boundary of the square region being investigated.
Check of the boundary y = 3 with x in [-1,3] f(x,3) = k(x) = x3 - 9x + 9 k'(x) = 3x2 - 9 = 0 if x = 31/2 and -31/2 -31/2 is not in [-1,3] k(31/2) = 9 - 6(31/2) which is greater than -3 k(-1) and k(3) would put us at corners of the square already evaluated above.
Thus the minimum value of the function over the region is -3 and occurs at (-1,-1,-3) and the maximum value of the function over the region is 37 and occurs at (3,-1,37).
Here is a DPGraph Picture of the surface f(x,y) = (x3 - 3xy + y2)/10 and the planes z = -0.3 and z = 3.7 graphed over the square region of the xy-coordinate plane described by the x-interval [-1,3] and y-interval [-1,3].
Section 13.8 #25 DPGraphpicture DPGraphpicture including the horizontal tangent planes (tilt the graph to see them both clearly) Quicktime movie showing many views Click on the Maple picture to see a different view. The blue traces intersect at (0,0,0) and the red traces intersect at (1,1,-1).
A MAPLE EXAMPLE WHERE THE COMPUTATION IN APPLYING THE SECOND PARTIALS TEST WOULD BE INTENSIVE
Section 13.8 #17 DPGraph Picture
EXAMPLES WHERE THE SECOND PARTIALS TEST FAILS
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