12.7 Extreme Values

# 12.7 Extreme Values - 1 2 .7 .  E x t r e m e  V a l...

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Unformatted text preview: 1 2 .7 .  E x t r e m e  V a l u e s   R 2      Local extrema occur at critical points where  f ′ ( x ) = 0  or  f ′ ( x )  is undefined.  Tests  1. First derivative test        2. Second derivative test          R3     , Local extrema occur at critical points where  ∇f ( xy ) = 0  or  ∇f ( x, y )   is undefined.    Note:  when  ∇f ( x, y ) = 0 ,  fx = 0  and fy = 0 simultaneously.    A saddle point is a point (a, b, f(a, b)) whose tangent plane is  horizontal ( ∇f ( x, y ) = 0 ) and f has both higher and lower  function values on any region containing (a, b).      Local extrema = local maxima, local minima, and saddle points.    Second Partials Test  Let z = f(x, y) and suppose  ∇f ( a, b) = 0 .  Define D(a, b), “the  discriminant of f”, as follows:  D(a, b) = fxx(a, b)fyy(a, b) – (fxy(a, b))2    1. If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum  at (a, b)  2. If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum  at (a, b)  3. If D(a, b) < 0, then f has a saddle point at (a, b)  4. If D(a, b) = 0, then it’s inconclusive    Ex.  f(x, y) = x4 + y3 + 32x – 9y                  Ex. f(x, y) = xy2 – 6x2 – 3y2                          Do:  Find all local extrema for f(x, y) = 6x2 ‐ 2x3 + 3y2 + 6xy                Absolute Maxima and Minima  If f is a continuous function on a closed and bounded region R,  then f has an absolute maximum and absolute minimum.  The  absolute maximum and minimum will always be at a critical  point or on the boundary.                                                  R2             1.0 0.5 -6 -4 -2 2 4 6 - 0.5 - 1.0                                R3                 Steps:  1. 2. 3. 4. Find the critical points and list them.  List the end points or corners  Find the critical points on the boundaries  Find the function values:  the largest is the absolute  maximum and the smallest is the absolute minimum.    Ex.  Find the absolute maximum and minimum for   f(x, y) = x4 + y3 + 32x – 9y on the region ‐2 ≤ x ≤ 0 and ‐2 ≤ y ≤ 0                                            Ex.  f(x, y) = xy2 – 6x2 – 3y2 on the triangular region with  vertices (1, 1), (‐1, 1), and (‐1, ‐3).                                                            Ex.  f(x, y) = xy2 – 6x2 – 3y2 on the region bounded by y = x2 and  y = 1.                                                            Ex.  f(x, y) = x + y2 on the region bounded by x2 + y2 ≤ 1.                                            ...
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## This note was uploaded on 10/02/2011 for the course AERO 1234 at Virginia Tech.

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