CALCULUS 4 QUIZ #3 REVIEW Part 2 / SPRING 09

Size: px

Start display at page:

Download "CALCULUS 4 QUIZ #3 REVIEW Part 2 / SPRING 09"

Kristina Sherman
5 years ago
Views:

1 CACUUS QUIZ #3 REVIEW Part / SPRING 09 (.) Determine the following about maima & minima of functions of variables. (a.) Complete the square for f( ) = + and locate all absolute maima & minima.. ( ) ( ) f ( ) = f ( ) = ( + ) ( ) The absolute maimum occurs at ( ). There is no absolute minimum. Absolute Maimum Point z Page of 5

2 (b.) ocate all relative maima minima & saddle points for f ( ) = 3. f = 3 = 0 f = = 0 = 3 ( ) = = 0 Here are the critical points: ( 0 0 0) 6 3. f = 6 f = f = D ( ) = f ( ) f ( ) f ( ) D ( ) = ( 6 ) ( ) ( ) = D0 ( 0) = < 0 Therefore ( 0 0 0) is a saddle point. D 6 = > 0 Therefore point. 6 3 is a ma. or min. Since f 6 < is a maimum. Page of 5

3 Saddle & Relative Maimum in close proimit z (c.) ocate all relative maima minima & saddle points for f ( ) + = +. d d f = = 0 d d f = = 0 3 = = = But the signs of "" and "" must be the same. Page 3 of 5

4 Accordingl here are the critical points: ( ) ( ). f = + 3 f = + 3 f = D ( ) = f ( ) f ( ) f ( ) D ( ) = D( ) = 3 > 0 D ( ) = 3 > 0 ( 3) Accordingl both points: ( ) ( ) must be relative maimums or minimums. Since f ( ) = 6 > 0 ( ) is a relative minimum. Since f ( ) = 6 > 0 ( ) is a relative minimum. Page of 5

5 Relative Minimum Points z (d.) Show that the second partials test provides no information about the critical points of the function f( ) = ; classif all critical points as relative maima relative minima or saddle points. f = 3 = 0 f = 3 = 0 Critical point is ( 0 0 0). f = f = f = 0 Page 5 of 5

6 D ( ) = f ( ) f ( ) f ( ) ( ) ( ) D ( ) = ( 0) = D0 ( 0) = 0 Since D( 0 0) = 0 the nd partials test tells us nothing about the nature of the critical point ( 0 0 0). z = This surface is a hperbolic paraboloid. It has a relative maimum at the point along the -ais and a relative minimum along the -ais. evidentl the point is a saddle point ( 0 0 0). Saddle (000) z Page 6 of 5

7 (e.) Find the absolute etrema of f( ) = 3 on the closed and bounded triangular region "R" with these vertices: ( 0 0) ( 0 ) ( 5 0). f = = 0 f = 3 = 0 Critical Point: ( 3 3) f ( ) = 0 f ( ) = 0 f ( ) = f ( 3 ) = 0 f ( 3 ) = 0 f ( 3 ) = In this case D < 0. This critical point is thus a Saddle Point. Also this point lies within the bounded region. Here is wh. R:Triangular Region Page 7 of 5

8 This is the equation of the hpotenuse. ( ) = 5 3 ( ) = 5 = 8 5 > > 0 Thus the point lies below the hpoteuse and above the base. Therefore it is an interior point and thus a candidate for an absolute etremum. Since " R " is "closed" and bounded there must eist both an Absolute Minimum & an Absolute Maimimum at Interior Points or on Boundar Points. Side # : = 0. f0 ( ) = 3 No Critical Points. Side # : = 0. f0 ( ) = No Critical Points. Side # 3 : = 5. f 5 = f 5 = 3 ( ) 5 Page 8 of 5

9 d f d 5 = 5 5 ( 3) d f d 5 = = 0 5 ( 5.) = = ( 3.375) =.3 8 Critical Point : ( ) Now we must evaluate f( ) at the three vertices. f00 ( ) = 0 f0 ( ) = f50 ( ) = 5 Evidentl the two vertices at ( 0 0 0) & ( 0 ) ield respectivel an Absolute Maimum & Absolute Minimum for " f ( ) " on " R ". Page 9 of 5

10 Ke Points in "R" z (f.) Find the absolute etrema of f( ) = on the closed and bounded region "R" where "R" is the disk +. Page 0 of 5

11 f() & Region: "R" z f ( ) = f = f = = 0 = 0 There is a horizontal tangent planes at the point: ( 0 0 0). Note that the critical point ( 0 0 0) is within the boundar of the circle. Page of 5

12 f = f ( 0 0) = f = f ( 0 0) = f = 0 f ( 0 0) = 0 D0 ( 0) = f ( 0 0) f ( 0 0) f ( 0 0) = < 0 Therefore the point: ( 0 0 0) is a saddle point. We break up the circle into an upper and lower branch and eam each branch separatel for etremals. u ( ) = l ( ) = f ( ) = v ( ) = f u ( ) = v ( ) = f u ( ) = = ( ) ( ) d d v ( ) = = 0 We get the same value for "" for the lower branch. Here are the points: ( 0 ) ( 0 ). Now we include the points: ( 0 ) ( 0 ) that we broke the circle into branches. Page of 5

13 Here are the candidates: ( 0 0 0) ( 0 ) ( 0 ) ( 0 ) and ( 0 ). Comparison of the "z" values of all 5 points reveals that the absolute maimum occurs at the points: ( 0 ) ( 0 ) and the absolute minimum occurs at the points: ( 0 ) ( 0 ). Absolute Etrermals in Region "R" z (g.) Show that among all parallelograms with perimeter a square with sides of length has maimum area. et the adjacent sides of the parallelogram be "a" and "b" with " θ" as the angle betwee those sides. Page 3 of 5

14 = A = a a b + b sin( θ) b = a ( ) = a Aa θ a sin( θ) a A = a sin( θ) = 0 θ A = a a cos( θ) = 0 a sin( θ) = 0 ( ) a cos θ = 0 The angle θ can not be zero because the resulting parallelogram would degenerate to a line and thus zero area. Thus the onl wa to satisf both equations simultaneousl is for their common factor a to vanish. a = 0 a = b = a = Thus the critical point a a a ( ) is a square. Page of 5

15 A aa = sin( θ) A aa ( a a) = sin π = < 0 Therefore out of all the paralleograms with a fied perimeter the square ields the maimum enclosed area. Page 5 of 5

MAXIMA & MINIMA The single-variable definitions and theorems relating to extermals can be extended to apply to multivariable calculus.

MAXIMA & MINIMA The single-variable definitions and theorems relating to etermals can be etended to appl to multivariable calculus. ( ) is a Relative Maimum if there ( ) such that ( ) f(, for all points