Finding the Extreme Values of Functions of Two Variables Calc IV Lab Karen Donnelly Saint Joseph's College

LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEocmVzdGFydEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIjpGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EsMC4yNzc3Nzc4ZW1GJy8lJ3JzcGFjZUdGTA== QyQ+SS1fRW52RXhwbGljaXRHNiJJJXRydWVHJSpwcm90ZWN0ZWRHISIi QyYtSSV3aXRoRzYiNiNJJnBsb3RzRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YlISIiLUYkNiNJKnBsb3R0b29sc0dGKEYr QyQtSSV3aXRoRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliRzYiNiNJK1JlYWxEb21haW5HRiUhIiI= Note that the above command "_EnvExplicit := true" makes sure that all roots of polynomials are shown as explicit solutions, and "with(RealDomain)" makes solve find only real solutions.

A continuous functions of two variables, z = f(x,y), defined on closed bounded region in the x-y plane will have absolute maximum and minimum values. Possible points (x,y) for maxima and minima of a function are: a) Boundary points of the region R on which the function is defined (if the region is bounded) (See Figure 12.63, and 12.71, in the book). b) Interior points (x,y) where the partial derivatives LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRInhGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ== and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRInlGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ== are both 0 or at least one fails to exist. These are called the critical points). To assist in determining whether the critical points are local (relative) maxima or local (relative) minima (or neither -- saddle points), we can use the Second Derivative Test (provided the first and second derivatives exist): (See text, page 1023 bottom): Assume that (a,b) is a critical point of f with first order partial derivatives LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRInhGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ==(a,b) and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRInlGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ==(a,b) both = 0. Define the discriminant of f at (a,b) to be: 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, then 1. If LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= > 0 and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JJW1zdWJHRiQ2JS1GLDYlUSJmRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUYjNiQtRiw2JVEjeHhGJ0Y3RjovRjtRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRictSSNtb0dGJDYtUTAmQXBwbHlGdW5jdGlvbjtGJ0ZCLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZNLyUpc3RyZXRjaHlHRk0vJSpzeW1tZXRyaWNHRk0vJShsYXJnZW9wR0ZNLyUubW92YWJsZWxpbWl0c0dGTS8lJ2FjY2VudEdGTS8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRmZuLUkobWZlbmNlZEdGJDYkLUYjNiYtRiw2JVEiYUYnRjdGOi1GSDYtUSIsRidGQkZLL0ZPRjlGUEZSRlRGVkZYRlovRmhuUSwwLjMzMzMzMzNlbUYnLUYsNiVRImJGJ0Y3RjpGQkZCRkJGK0ZC < 0, ( a,b), is a local maximum 2. If LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= > 0 and 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 > 0, ( a,b) is a local minimum. 3. If LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= < 0, then (a,b) is a saddle point. 4. If LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= = 0, the test is inconclusive -- it could be anything. To find the absolute maximum and absolute minimum, compare the boundary point values (if the function is defined on a bounded region) and the identified local maxs and mins to see which values are largest and smallest. In this lab these techniques are applied to finding extreme values of a specific functions of two variables.

1. Consider the function f defined as: 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. PkkiZkc2ImYqNiRJInhHRiRJInlHRiRGJDYkSSlvcGVyYXRvckdGJEkmYXJyb3dHRiRGJCwkKig5JCIiIjklRi8tSSRleHBHRiQ2IywmKiRGLiIiIyMhIiJGNiokRjBGNkY3Ri8iIzVGJEYkRiQ= 2. This function is defined for all points (x,y) so there are no boundary points. We plot the graph to see get in indication of the extreme values. If you switch to the contour view and look down from the top you can get a good idea of the critical points(x,y). QyQ+SSJwRzYiLUkncGxvdDNkR0YlNiYtSSJmR0YlNiRJInhHRiVJInlHRiUvRiw7ISIkIiIkL0YtRi8vSSVheGVzR0YlSSZCT1hFREdGJSIiIg==SSIlRzYi 3. This plot suggests that there appears to be an extreme value in each quadrant. There appears to be minima in the second and fourth quadrants and maxima in the first and third quadrants. Another way of seeing approximately what the critical points are is to use the contourplot function which gives some of the level curves: . LUkld2l0aEc2JCUqcHJvdGVjdGVkR0koX3N5c2xpYkc2IjYjSSZwbG90c0dGJA== LUksY29udG91cnBsb3RHNiI2Ji1JImZHRiQ2JEkieEdGJEkieUdGJC9GKTshIiQiIiQvRipGLC9JKWNvbnRvdXJzR0YkIiM6 4. Now apply the second derivative test. First calculate the first-order partial derivatives. The factor command helps to simplify the answer so that we can determine where it is equal to 0. PkkjZnhHNiItSSdmYWN0b3JHRiQ2Iy1JJWRpZmZHJSpwcm90ZWN0ZWRHNiQtSSJmR0YkNiRJInhHRiRJInlHRiRGLw== PkkjZnlHNiItSSdmYWN0b3JHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHRiQ2Iy1JJWRpZmZHRig2JC1JImZHRiQ2JEkieEdGJEkieUdGJEYy 5. There are five critical points (i.e. points where LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRInhGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ== = LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRInlGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ== = 0 ). They are (0,0), (1,1), (1,-1), (-1,1), and (-1,-1). We can see this by inspection above or making Maple do the dirty work. LUkmc29sdmVHNiI2JDwkL0kjZnhHRiQiIiEvSSNmeUdGJEYpPCRJInhHRiRJInlHRiQ= 6. Just by looking at the 3-D plot of the function, we would conclude that that the points (1,1) and (-1,-1) are maxima, and points (-1,1), (1,-1) are minima and (0,0) is a saddle point. We can confirm this observation by applying the second derivative test. The following commands compute (and then factor) the second order partial derivatives. PkkkZnh4RzYiLUknZmFjdG9yRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YkNiMtSSVkaWZmR0YoNiRJI2Z4R0YkSSJ4R0Yk PkkkZnl5RzYiLUknZmFjdG9yRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YkNiMtSSVkaWZmR0YoNiRJI2Z5R0YkSSJ5R0Yk PkkkZnh5RzYiLUknZmFjdG9yRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YkNiMtSSVkaWZmR0YoNiRJI2Z4R0YkSSJ5R0Yk 7. The following computes the discriminant LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= . PkkiZEc2IiwmKiZJJGZ4eEdGJCIiIkkkZnl5R0YkRihGKC1JIl5HNiQlKnByb3RlY3RlZEcvSSttb2R1bGVuYW1lR0YkSStSZWFsRG9tYWluRzYkRi1JKF9zeXNsaWJHRiQ2JEkkZnh5R0YkIiIjISIi PkkiZEc2Ii1JJ2ZhY3Rvckc2JCUqcHJvdGVjdGVkR0koX3N5c2xpYkdGJDYjRiM= 8. Now we apply the second derivative test to each of the critical points, starting with the critical point (0,0): LUklc3Vic0clKnByb3RlY3RlZEc2JS9JInhHNiIiIiEvSSJ5R0YoRilJImRHRig= LUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= Thus LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= < 0 at the point (0,0) and this point is a saddle point. LUkiZkc2IjYkIiIhRiY= 9. The point (1,1): Check the values of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRI3h4RidGMkY1L0Y2USdub3JtYWxGJy8lL3N1YnNjcmlwdHNoaWZ0R1EiMEYnRj0= and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= at the point (1,1). QyQtSSVzdWJzRyUqcHJvdGVjdGVkRzYlL0kieEc2IiIiIi9JInlHRilGKkkiZEdGKUYqLUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= QyQtSSVzdWJzRyUqcHJvdGVjdGVkRzYlL0kieEc2IiIiIi9JInlHRilGKkkkZnh4R0YpRio=LUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= Since LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= > 0 , and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRI3h4RidGMkY1L0Y2USdub3JtYWxGJy8lL3N1YnNjcmlwdHNoaWZ0R1EiMEYnRj0= < 0 at (1,1), (1,1) is a point where a local maximum occurs. QyUtSSJmRzYiNiQiIiJGJ0YnLUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHRiU= 10. The critical point (-1,1): QyQtSSVzdWJzRyUqcHJvdGVjdGVkRzYlL0kieEc2IiEiIi9JInlHRikiIiJJImRHRilGLQ==LUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= QyQtSSVzdWJzRyUqcHJvdGVjdGVkRzYlL0kieEc2IiEiIi9JInlHRikiIiJJJGZ4eEdGKUYtLUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= Thus f(-1,1) is a local minimum QyUtSSJmRzYiNiQhIiIiIiJGKC1JJmV2YWxmRyUqcHJvdGVjdGVkRzYjSSIlR0Yl 11. Finally, for (-1,-1) and (1,-1) -- getting it over in a hurry: QyQtSSVzdWJzRyUqcHJvdGVjdGVkRzYlL0kieEc2IiEiIi9JInlHRilGKjclSSJkR0YpSSRmeHhHRiktSSJmR0YpNiRGKEYsIiIiLUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= QyQtSSVzdWJzRyUqcHJvdGVjdGVkRzYlL0kieEc2IiIiIi9JInlHRikhIiI3JUkiZEdGKUkkZnh4R0YpLUkiZkdGKTYkRihGLEYqLUkmZXZhbGZHJSpwcm90ZWN0ZWRHNiNJIiVHNiI= Therefore f(-1,-1) is a relative maximum and f(1,-1) is a relative minimum. Summary: (a,b) LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= 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 Conclusion 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 (0,0) Negative 0 Saddle Point 0 (1,1) Positive Negative Relative Maximum .368 (-1,1) Positive Positive Relative Minimum -.368 (-1,-1) Positive Negative Relative Maximum .368 (1,-1) Positive Positive Relative Minimum -.368 Below we plot points at each of the identified critical points in different colors to see graphically that our test results are correct: QyQ+SSNwMUc2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSIiIUYqLUkiZkdGJTYkRipGKi9JJmNvbG9yR0YlSSZncmVlbkdGJSEiIg== QyQ+SSNwMkc2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSIiIkYqLUkiZkdGJTYkRipGKi9JJmNvbG9yR0YlSSRyZWRHRiUhIiI= QyQ+SSNwM0c2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSEiIiIiIi1JImZHRiU2JEYqRisvSSZjb2xvckdGJUklYmx1ZUdGJUYq QyQ+SSNwNEc2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSEiIkYqLUkiZkdGJTYkRipGKi9JJmNvbG9yR0YlSSZibGFja0dGJUYq QyQ+SSNwNUc2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSIiIiEiIi1JImZHRiU2JEYqRisvSSZjb2xvckdGJUkobWFnZW50YUdGJUYr LUkoZGlzcGxheUc2IjYqSSJwR0YkSSNwMUdGJEkjcDJHRiRJI3AzR0YkSSNwNEdGJEkjcDVHRiQvSSZzdHlsZUdGJEkscGF0Y2hub2dyaWRHRiQvSS10cmFuc3BhcmVuY3lHRiQkIiImISIi

Absolute extremum of a function of two variables defined on a closed region will occur either at critical points or along the boundary of the region. Thus we must find critical points and compare their values against the maximum and minimum values along points on the boundary of the region. Consider the function 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. Find its absolute extrema over the triangular region bounded by y = 0, x = 4 and y = 2x. PkkiZkc2ImYqNiRJInhHRiRJInlHRiRGJDYkSSlvcGVyYXRvckdGJEkmYXJyb3dHRiRGJCwoKiY5JCIiIjklRi9GL0YuISIjRjAhIiRGJEYkRiQ= Plotting note: Notice below the way the triangular region can be defined as the plotting domain by using a suitable variable end-point for the y-interval. (When needed, both end-points of the y-interval can be treated in this way.) QyY+SSJwRzYiLUkncGxvdDNkRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YlNigtSSJmR0YlNiRJInhHRiVJInlHRiUvRi87IiIhIiIlL0YwO0YzLCRGLyIiIy9JJXZpZXdHRiU3JUYyO0YzIiIpOyEiKUY4L0klYXhlc0dGJUkmYm94ZWRHRiUvSS10cmFuc3BhcmVuY3lHRiUkIiImISIiRkdGJCIiIg== Rotate the graph above so that you are looking directly down into the x,y-plane to see the triangular region. 4. Now apply the second derivative test. First calculate the first-order partial derivatives. The factor command helps to simplify the answer so that we can determine where it is equal to 0. PkkjZnhHNiItSSVkaWZmRyUqcHJvdGVjdGVkRzYkLUkiZkdGJDYkSSJ4R0YkSSJ5R0YkRiw= PkkjZnlHNiItSSVkaWZmRyUqcHJvdGVjdGVkRzYkLUkiZkdGJDYkSSJ4R0YkSSJ5R0YkRi0= Obviously the critical point is 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 -- But we can ask Maple anyway: LUkmc29sdmVHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2JDwkL0kjZnhHRiciIiEvSSNmeUdGJ0YsPCRJInhHRidJInlHRic= PkkkZnh4RzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZnhHRiRJInhHRiQ= PkkkZnh5RzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZnhHRiRJInlHRiQ= PkkkZnl5RzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZnlHRiRJInlHRiQ= PkkiZEc2IiwmKiZJJGZ4eEdGJCIiIkkkZnl5R0YkRihGKC1JIl5HNiQlKnByb3RlY3RlZEcvSSttb2R1bGVuYW1lR0YkSStSZWFsRG9tYWluRzYkRi1JKF9zeXNsaWJHRiQ2JEkkZnh5R0YkIiIjISIi We don't need to substitute in values for x and y since the discriminant is a constant. Since the discriminant is negative, this is a saddle point so we can ignore it as a potential max or minimum. We can also see this from the graph: QyQ+SSNTUEc2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSIiJCIiIy1JImZHRiU2JEYqRisvSSZjb2xvckdGJUkkcmVkR0YlISIi LUkoZGlzcGxheUc2IjYmSSJwR0YkSSNTUEdGJC9JJWF4ZXNHRiRJJGJveEdGJC9JJnN0eWxlR0YkSS1wYXRjaGNvbnRvdXJHRiQ= Now we explore the boundary of the triangle, given by the three lines 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. Inspection of the graph leads us to believe: Absolute maximum value of f is 0, attained at (0,0) and at (4,8), Absolute minimum value is -8, attained at (4,0) and perhaps at a second point somewhere near the middle of the border line 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. Check out the values of the function at the first three mentioned points: NiUtSSJmRzYiNiQiIiFGJy1GJDYkIiIlIiIpLUYkNiRGKkYn a) Check the value of f(x,y) along the border line x = 4: LUklc3Vic0clKnByb3RlY3RlZEc2JC9JInhHNiIiIiUtSSJmR0YoNiRGJ0kieUdGKA== Clearly this since y varies between 0 and 8 in this triangle, the largest value of the function 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 is 0 (when y is 8), and the smallest is -8 (when y is equal to 0). b) Check the value of the function 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along the border line y = 0: LUklc3Vic0clKnByb3RlY3RlZEc2JC9JInlHNiIiIiEtSSJmR0YoNiRJInhHRihGJw== Along this line y = 0, the maximum of 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 occurs at (0,0) and has value 0; The minimum occurs at (4,0) and has value -8. c) Along the line 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: Substituting this for y = 2x in the function f, we obtain PkkiaEc2Ii1JJXN1YnNHJSpwcm90ZWN0ZWRHNiQvSSJ5R0YkLCRJInhHRiQiIiMtSSJmR0YkNiRGLEYq Applying the second derivative test to this ordinary function of a single variable: PkkjZGhHNiItSSVkaWZmRyUqcHJvdGVjdGVkRzYkSSJoR0YkSSJ4R0YkJSFH PkkkZGgyRzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZGhHRiRJInhHRiQ= LUkmc29sdmVHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2JC9JI2RoR0YnIiIhSSJ4R0Yn This critical point of x = 2, y = 4, for h corresponds to a relative min along the borderL0kieUc2IiwkSSJ4R0YkIiIj of our function LUkiZkc2IjYkIiIjIiIl Thus this point (2,4,-8) "ties" as an absolute minimum with the border point (4,0,-8). We mark this point with a green point: QyQ+SSNwMUc2Ii1JLHBvaW50cGxvdDNkR0YlNiQ3JSIiIyIiJSEiKS9JJmNvbG9yR0YlSSZncmVlbkdGJSEiIg== LUkoZGlzcGxheUc2IjYoSSJwR0YkSSNTUEdGJEkjcDFHRiQvSSZzdHlsZUdGJEktcGF0Y2hjb250b3VyR0YkL0klYXhlc0dGJEkkYm94R0YkL0ktdHJhbnNwYXJlbmN5R0YkJCIiJiEiIg== Summary: Critical Point: (a,b) LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRicvRjNRJ25vcm1hbEYn 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 Conclusion 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 (2,3) Negative 0 Saddle Point -6 (Ignore this point) Extrema on the border defined by x = 2; y = 4; and y = 2x Absolute maximum of 0 at (x, y) = (0, 0) and (4, 8) Absolute minimum of -8 at (LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYnLUkjbWlHRiQ2I1EhRictRiM2J0YrLUYjNictRiw2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIixGJy9GOlEnbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRjgvJSlzdHJldGNoeUdGRC8lKnN5bW1ldHJpY0dGRC8lKGxhcmdlb3BHRkQvJS5tb3ZhYmxlbGltaXRzR0ZELyUnYWNjZW50R0ZELyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdRLDAuMzMzMzMzM2VtRictRiw2JVEieUYnRjZGOS8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRidGQEYrRlpGQEYrRlpGQA==) = (4, 0) and (2, 4)

Using the Maple techniques as above to employ the 2nd derivative test to find any relative extrema or saddle points for the function defined by LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzZdby1JI21vR0YkNi1RIn5GJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRjQvJSlzdHJldGNoeUdGNC8lKnN5bW1ldHJpY0dGNC8lKGxhcmdlb3BHRjQvJS5tb3ZhYmxlbGltaXRzR0Y0LyUnYWNjZW50R0Y0LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGQ0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRitGK0YrRistSSNtaUdGJDYlUSJmRicvJSdpdGFsaWNHUSV0cnVlRicvRjBRJ2l0YWxpY0YnLUkobWZlbmNlZEdGJDYkLUYjNictRkc2JVEieEYnRkpGTS1GLDYtUSIsRidGL0YyL0Y2RkxGN0Y5RjtGPUY/RkEvRkVRLDAuMzMzMzMzM2VtRictRkc2JVEieUYnRkpGTS8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRidGL0YvLUYsNi1RIj1GJ0YvRjJGNUY3RjlGO0Y9Rj8vRkJRLDAuMjc3Nzc3OGVtRicvRkVGYW8tSSVtc3VwR0YkNiVGVC1JI21uR0YkNiRRIjNGJ0YvLyUxc3VwZXJzY3JpcHRzaGlmdEdRIjBGJy1GLDYtUSgmbWludXM7RidGL0YyRjVGN0Y5RjtGPUY/L0ZCUSwwLjIyMjIyMjJlbUYnL0ZFRmFwRmZvRitGVEYrRmduLUYsNi1RIitGJ0YvRjJGNUY3RjlGO0Y9Rj9GYHBGYnAtRmRvNiVGZ25GZm9Gam9Gam5GLw== PkkiZkc2ImYqNiRJInhHRiRJInlHRiRGJDYkSSlvcGVyYXRvckdGJEkmYXJyb3dHRiRGJCwoKiQ5JCIiJCIiIiomRi5GMDklRjAhIiQqJEYyRi9GMEYkRiRGJA== QyQ+SSJwRzYiLUkncGxvdDNkRzYkJSpwcm90ZWN0ZWRHSShfc3lzbGliR0YlNigtSSJmR0YlNiRJInhHRiVJInlHRiUvRi87ISIkIiIkL0YwRjIvSSVheGVzR0YlSSdub3JtYWxHRikvSS10cmFuc3BhcmVuY3lHRiUkIiImISIiL0kmc3R5bGVHRiVJLXBhdGNoY29udG91ckdGJUY9SSIlRzYi PkkjZnhHNiItSSVkaWZmRyUqcHJvdGVjdGVkRzYkLUkiZkdGJDYkSSJ4R0YkSSJ5R0YkRiw= PkkjZnlHNiItSSVkaWZmRyUqcHJvdGVjdGVkRzYkLUkiZkdGJDYkSSJ4R0YkSSJ5R0YkRi0= LUkmc29sdmVHNiQlKnByb3RlY3RlZEdJKF9zeXNsaWJHNiI2JDwkL0kjZnhHRiciIiEvSSNmeUdGJ0YsPCRJInhHRidJInlHRic= PkkkZnh4RzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZnhHRiRJInhHRiQ= PkkkZnh5RzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZnhHRiRJInlHRiQ= PkkkZnl5RzYiLUklZGlmZkclKnByb3RlY3RlZEc2JEkjZnlHRiRJInlHRiQ= PkkiZEc2IiwmKiZJJGZ4eEdGJCIiIkkkZnl5R0YkRihGKC1JIl5HNiQlKnByb3RlY3RlZEcvSSttb2R1bGVuYW1lR0YkSStSZWFsRG9tYWluRzYkRi1JKF9zeXNsaWJHRiQ2JEkkZnh5R0YkIiIjISIi LUklc3Vic0clKnByb3RlY3RlZEc2JS9JInhHNiIiIiEvSSJ5R0YoRilJImRHRig= Thus LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= < 0 at the point (0,0) and this point is a saddle point. LUkiZkc2IjYkIiIhRiY= The point (1,1): Check the values of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRI3h4RidGMkY1L0Y2USdub3JtYWxGJy8lL3N1YnNjcmlwdHNoaWZ0R1EiMEYnRj0= and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= at the point (1,1). LUklc3Vic0clKnByb3RlY3RlZEc2JS9JInhHNiIiIiIvSSJ5R0YoRilJImRHRig= LUklc3Vic0clKnByb3RlY3RlZEc2JS9JInhHNiIiIiIvSSJ5R0YoRilJJGZ4eEdGKA== Since LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiZEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= > 0 , and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRI3h4RidGMkY1L0Y2USdub3JtYWxGJy8lL3N1YnNjcmlwdHNoaWZ0R1EiMEYnRj0= > 0 at (1,1), (1,1) is a point where a local minimum occurs. LUkiZkc2IjYkIiIiRiY= LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=

solve factor From the plots package pointplot3d