Notes for 6 1 and 6 2 Math 211 In these two sections we discuss methods by which to locate minimum and maximum points of functions z f x y in 3 dimensions The techniques used are virtually identical to those used to locate extrema points in 2 dimensions graphs of the form y f x To review let us do an example from singlevariable calculus Example Find the extrema value s of the function f x 2 x 3 3 x 2 12 x 2 Solution Extreme values occur when the first derivative f x 0 or is undefined e g a corner as in the absolute value graph The derivative in this case is f x 6 x 2 6 x 12 Set f x 0 factor and solve and we get 6 x 2 6 x 12 0 6 x 2 x 2 0 6 x 1 x 2 0 which gives two critical values for x x 1 and x 2 There are two common methods to determine the nature of these critical values The first derivative test looks for sign changes of f x between the critical values to infer min and max while the second derivative test looks for the concavity of the function to infer min or max We ll use the second derivative test here since we ll use its analog in the multivariable case The second derivative is f x 12 x 6 By evaluating the second derivative at each of the critical points and looking at the sign of the result we can determine whether we have min max or possibly nothing In this case f 1 18 and f 2 18 Therefore the function has a local minimum at x 2 since its second derivative value is positive which means the function opens up is concave up The function has a local max at x 1 since its second derivative value at that point is negative meaning the function is concave down The actual point values are 1 9 and 2 18 Note Not all functions have minimum or maximum values The function f x x 2 has a minimum value at 0 0 but no maximum while f x x 3 has a critical value at x 0 but neither a minimum nor maximum value here ASU Math Surgent Math 211 Multivariable Extrema For our purposes in sections 7 1 and 7 2 we will be looking at paraboloids which are the multivariable calculus analogs to normal parabolas in single variable calculus The usual equation of a paraboloid is f x y Ax 2 By 2 Cxy Dx Ey F where A F are coefficients Our goal is to determine the location of its vertex and to see whether it is a minimum point maximum point or possibly a saddle point We perform essentially the same steps in multivariable settings as with single variable problems We start by taking the first partial derivatives of f x y Remember f x represents the first partial derivative of f x y meaning the derivative of the function in the x direction directions parallel to the x axis Similarly for f y The problem here is that f x might equal zero at some point but f y may not or vice versa We need to see where both f x and f y equal zero simultaneously This means we will have to set up a simple 2 by 2 system to calculate the x and y values at which this happens Example Determine the critical point for the paraboloid f x y x 2 y 2 xy x 5 y 2 Solution find the first partial derivatives and form a system by setting both equations to zero then solving using elimination or substitution f x 2x y 1 2x y 1 0 2x y 1 x 1 f y 2y x 5 2y x 5 0 x 2y 5 y 3 The z coordinate can be found by evaluating the function at the critical values for x and y z f 1 3 1 2 3 2 1 3 1 5 3 2 5 Hence the critical point is 1 3 5 Is it a minimum maximum or neither The second derivative test for surfaces is analogous to the case for single variable calculus but with some added steps First determine all second derivatives f xx 2 f xy 1 f yy 2 f yx 1 Recall that for smooth continuous well behaved functions like paraboloids the mixed partial derivatives should always be equal as a check of your work Next we calculate the value of the second derivative call it at the critical point by the following equation f xx f yy f xy f yx This can also be written f xx f yy f xy 2 Do you see why Important For paraboloids the second derivatives will always be constants For other types of surfaces the second derivatives will likely have variables in which you need to evaluate at the critical values for x and y ASU Math Surgent Math 211 Multivariable Extrema In the example from the previous page we get 2 2 1 1 3 As in the single variable case we usually don t care about the value of just its sign However we need to follow the following rules Case I 0 and f xx 0 This implies we have a minimum at the critical point For the example we are working on the point 1 3 5 is the minimum point of this surface Case II 0 and f xx 0 The critical point is a maximum Case III 0 The critical point is a saddle point neither a minimum nor maximum The surface graph of the example function General observations Let f x y Ax 2 By 2 Cxy Dx Ey F represent the generic paraboloid The only coefficients that have a direct effect on the nature of the graph are A B and C Note that f xx 2 A f yy 2 B f xy f yx C If both A and B are positive then the vertex is probably a minimum provided the value of C isn t too large Similarly if both A and B are negative the vertex is probably a maximum again provided C isn t too large If C is large such that 2 A 2 B C 2 is negative then the vertex is a saddle Also if A and B have different signs then the vertex is always a saddle point For an example of a saddle point let s re use our same sample function but increase the C value in front of the xy term to a larger value f x y x 2 y 2 3 xy x 5 y 2 Note that the second derivative value for will now be negative For a graph see the next page ASU Math Surgent Math 211 Multivariable Extrema Graph of f x y x 2 y 2 3 xy x 5 y 2 ASU Math Surgent Math 211 Multivariable Extrema