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It was the best of times, it was the worst of times Copyright 2008 by Evans M. Harrell II.About the test  Median was 80.  As it is written:About the test  Percentiles:  90th: 97  75th: 89  50th: 80  25th: 63 (Seek help.) Range: 32 to 100Absolute max and min  f(x0) ≥ f(x) for all x in D.  THE THEOREM: If f is a _continuous_ function on a _closed, bounded_ set D, then f takes on an absolute maximum on D. Also an absolute minimum.Local max or min  Takes place at a critical point  Gradient = 0 (all components) OR  Gradient undefinedLudwig Otto Hesse – the Determinator Second derivative test?Second derivative test  D = det (H) = fxx fyy - (fxy)2. If at a crit. pt, and  D < 0, then SADDLE.  D > 0, then LOCAL MAX OR MIN. Check fxx or fyy to determine which: fxx > 0 ⇒ min, < 0 ⇒ maxExamples - Find and classify critical points  4 x3 + y2 - 12 x2 - 36 x  x y e-2xy  x3 + (x - y)2Examples - Find and classify critical points  4 x3 + y2 - 12 x2 - 36 x  (3, 0) is a local minimum (Hessian matrix has positives on diagonal and positive determinant.)  (-1,0) is a saddle. (Hessian matrix has negative determinant.)Example - x y e-2xy Note: y-2xy2 = y(1 – 2xy)A funky exampleA funky exampleExtra credit contest due next week Due Thursday, 9 October. This contest has to do with the "funky example.” Find a formula for a saddle like that one. What do the gradient and Hessian matrix tell you about the funky saddle point at the origin x = y = 0? Carefully discuss the tangent planes and the (3D) normal vectors at points near the origin.Second derivative test? Guess the theorem for 3D!Second derivative test  Find the eigenvalues of the Hessian matrix at a critical point.  If all of them are < 0, then LOCAL MAX  If all of them are < 0, then LOCAL MAX  Is some are < 0 and some are > 0, then the critical point is something like a saddle.  If one or more eigenvalues are 0, inconclusive.Examples  f(x,y,z) = x2 + 4 y2 – 8 y + 3 z2 + xy – yz +2  f(x,y,z) = x2 - 4 y2 – 8 y + 3 z2 + xy – yz +2  f(x,y,z) = 7 - x2 - 4 y2 – 8 y - 3 z2 + xy – yzAbsolute max and min  f(x0) ≥ f(x) for all x in D.  THE THEOREM: If f is a _continuous_ function on a _closed, bounded_ set D, then f takes on an absolute maximum on D. Also an absolute minimum.Absolute max and min  Here is the algorithm for a closed, bounded region and continuous function:  Calculate gradient  List all critical points  Optional: Use Hessian test to eliminate some candidates.  Also check the boundary points  (more on this subject in future lecture) How can you minimize the amount of wood needed to make a rectangular box, with no top, holding 1 cubic meter? A word problem – “modeling”A word problem – “modeling”  How can you minimize the amount of wood needed to make a rectangular box, with no top, holding 1 cubic meter?  The objective function is xy+2 xz + 2 yz.  The constraints are: 0 ≤ x, y, z, and xyz=1.  Use one constraint to eliminate one variable.One option: Use the constraint to reduce the dimension. Substitute z = 1/xy to get an objective function of the form f(x,y) = x y + 2/x + 2/y . Although this region is unbounded, we intuitively know that the box will have finite dimensions. As for the constraints 0 ≤ x,y,z, we also know that on the boundary: {x=0} or {y=0} or {z=0}, the box would have 0 volume. The good values (x,y) are in the interior, and therefore are a critical point.One option: Use the constraint to reduce the dimension. Substitute z = 1/xy to get an objective function of the form f(x,y) = x y + 2/x + 2/y . The gradient is ∇f(x,y) = (y - 2/x2)i + (x - 2/y2)j  Solving gives x = y = 21/3 as the only critical point.  z is then found to be 2-2/3.  The amount of wood needed is 3 • 22/3 = 4.7622… mExample  Let f(x,y) = x2 + y2 + x + y on the closed unit disk. Find its absolute max and min.  Only 1 critical pt: 2x+1= 0 = 2y+1 ⇒ x=y=-1/2.Example  Note that f(x,y) = x2 + y2 + x + y can also be written as (x+½)2 + (x+½)2 - ½ (Complete the square.)Example  Let f(x,y) = x2 + y2 + x + y on the closed unit disk. Find its absolute max and min.  Only 1 critical pt: 2x+1= 0 = 2y+1 ⇒ x=y=-1/2.! On the boundary write x = cos t, y = sin t  f becomes 1+cos t+sin t and we are back in KG Calculus.Example  Let f(x,y) = x2 + y2 + x + y on the closed unit disk. Find its absolute max and min.  The critical points of 1+cos t+sin t occur when cos t = sin t, I.e., x = y.  The boundary “candidates” for Max and Minnie are therefore x = y = 2-1/2 and x = y = - 2-1/2.Example  Plug the three candidates into f(x,y) = x2 + y2 + x + y :  The f(-1/2, -1/2) = -1/2 = -0.5  f(2-1/2,2-1/2) = 1+ 21/2 = 2.414214…  f(-2-1/2,-2-1/2) = 1- 21/2 = -0.414214… Max! Minnie! This candidate is a loser!Constraints, regions, and side conditions  These are all pretty much the same!  We can think of it as a region when x,y,z are position variables, even when the physical meaning is entirely different:  Temperature, pressure, volume  Cost of various items,


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GT MATH 2401 - Lecture Notes

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