( ) 2 3x=0 3x(x 3 1)=0 x=0 x=1

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1 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values. (a) First we compute D(,)= f (,) f (,) [ f (,)] =(4)() () =7. Since D(,)>0 and f (,)>0, f has a local minimum at (,) b the Second Derivatives Test. (b) D(,)= f (,) f (,) [ f (,)] =(4)() () =. Since D(,)<0, f has a saddle point at (,) b the Second Derivatives Test.. (a) D=g (0,)g (0,) [g (0,)] =() (6) = 7. Since D<0, g has a saddle point at (0,) b the Second Derivatives Test. (b) D=g (0,)g (0,) [g (0,)] =( 8) () =4. Since D>0 and g (0,)<0, g has a local maimum at (0,) b the Second Derivatives Test. (c) D=g (0,)g (0,) [g (0,)] =(4)(9) (6) =0. In this case the Second Derivatives Test gives no information about g at the point (0,).. In the figure, a point at approimatel (,) is enclosed b level curves which are oval in shape and indicate that as we move awa from the point in an direction the values of f are increasing. Hence we would epect a local minimum at or near (,). The level curves near ( 0,0) resemble hperbolas, and as we move awa from the origin, the values of f increase in some directions and decrease in others, so we would epect to find a saddle point there. To verif our predictions, we have f (,)=4+ + f (,)=, f (,)=. We have critical points where these partial derivatives are equal to 0 : =0, =0. Substituting = =0 ( )=0 =0 = from the first equation into the second equation gives or. Then we have two critical points, 0,0 and. The second partial derivatives are, f (,)=, and f (,)=6, so D(,)= f (,) f (,) [ f (,)] =(6)(6) (,) f (,)=6 =6 9 D(0,0)=6(0)(0) 9= 9. Then, and D(,)=6()() 9=7. Since D(0,0)<0, f has a saddle point at (0,0) b the Second Derivatives Test. Since D(,)>0 and f (,)>0, f has a local minimum at (,). 4. In the figure, points at approimatel (,) and (, ) are enclosed b oval shaped level curves which indicate that as we move awa from either point in an direction, the values of f are increasing. Hence we would epect local minima at or near (, ). Similarl, the point (,0) appears to be enclosed b oval shaped level curves which indicate that as we move awa from the point in an direction the values of f are decreasing, so we should have a local maimum there. We also show hperbola shaped level curves near the points (,0), (,), and (, ). The values of f increase along some paths leaving these points and decrease in others, so we should have a saddle

2 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values point at each of these points. To confirm our predictions, we have f (,)= + 4 f (,)=, f (,)= 4+4. Setting these partial derivatives equal to 0, we have =0 = and 4+4 =0 ( )=0 =0,. So our critical points are (,0), (, ). The second partial derivatives are f (,)= 6, f (,)=0, and f (,)= 4, so D(,)= f (,) f (,) [ f (,)] =( 6) 4 Derivatives Test to classif the 6 critical points: Critical Point D f Conclusion (0) = 7 +4 (,0) 4 6 D>0, f <0 f has a local maimum at,0 (,) 48 D<0 f has a saddle point at (,) (, ) 48 D<0 f has a saddle point at (, ) (,0) 4 D<0 f has a saddle point at (,0), has a local minimum at 48 6 D>0, f >0 f, (, ) 48 6 D>0, f >0 f has a local minimum at,. We use the Second 5. f (,)= f =, f =4 8, f =, f =0, f = 8. Then f =0 and f =0 impl = and =, and the onl critical point is,. D(,)= f f f, ( f ) = ( ) 8 0 =6 D, = is a local maimum b the Second Derivatives Test., and since =6>0 and f, = <0, 6. f (,)= + 8 f = +4, f = 8, f =6+4, f =, f =0. Then f =0 implies =, and substitution into f =0 gives D(, 4)=( 4)(0) = 44<0 (, 4) +48=0 = 4. Thus, the onl critical point is, 4., so is a saddle point.

3 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values 7. f (,)= f =4 4, f =4 4, f =, f = 4, f =. Then f =0 implies =, and substitution into f =0 = gives 9 =0 ( 8 )=0 =0 or =. Thus the critical points are ( 0,0), (,), and (, ). Now D(0,0)=0 0 ( 4) = 6<0, so ( 0,0) is a saddle point. D(,)=()() ( 4) >0 and f (,)=>0, so f (,)=0 is a local minimum. D(, )=()() ( 4) >0 and f =(, )=>0, so f (, )=0 is also a local minimum. f (,)=e 4 f = e 4 f =(4 )e 4 f =(4 )e 4 8.,,, f = (4 )e 4 f =(4 6+4)e 4,. Then f =0 and f =0 implies =0 and =, so the onl critical point is (0,). D(0,)=( e 4 )( e 4 ) 0 =4e 8 >0 and f (0,)= e 4 <0, so f (0,)=e 4 is a local maimum.

4 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values 9. f (,)=(+)(+)=++ + f =++, f =+ +, f =, f =+, f =. Then f =0 implies ++ =0 and f =0 implies + +=0. Subtracting the second equation from the first gives =0 =, but if = then ++ =0 + =0 which has no real solution. If = then ++ =0 =0 =, so critical points are (, ) and,. and, so and, are saddle points. D(, )=() 0<0 D(,)=() 0<0 (,) 0. f (,)= f =6 + +0, f =+, f =+0, f =+, f =. Then 5 f =0 implies =0 or =. Substituting into f =0 gives the critical points 0,0,,0,,. Now and, so is a local minimum. Also D(0,0)=0>0 f (0,0)=0>0 f (0,0)=0 f while 5,0 <0 D 5,0 >0 D(, )<0 f 5 5,, and. Hence,0 = is a local maimum 7, are saddle points.. f (,)=+ f =, f =, f = f =, f =. Then f =0 and f =0 implies D(, 0 0 ) = so the critical points are all points of the form,. But =4 4=0 so the Second 0 0 ( ) (,) Derivatives Test gives no information. However + = and for all, with equalit if and onl if =. Thus f, are local maima. 0 0 = 4

5 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values. f (,)=( ) f =, f =, f =, f =, f =. Then f =0 implies =0 or =. Substituting =0 into f =0 gives =0 or = and substituting = into f =0 gives =0 so =0 or. Thus the critical points are 0,0, (,0), ( 0,) and,. D ( 0,0)=D (,0)=D ( 0,)= D, = f, while and = <0. Thus 0,0, and ( 0,) are saddle points, and f, = is a local maimum. 7 (,0). f (,)=e cos f =e cos, f = e sin. Now f =0 implies cos =0 or = +n for n an integer. But sin +n 0, so there are no critical points. f (,)= + + f = 4., f =, f =+6 4, f =+6 4, f =4. Then f =0 implies 4 =0 or 4 = or = 4. Note that neither nor can be zero. Now f =0 implies 4 =0, and with = 4 this implies 6 =0 or 6 =. Thus = and 5

6 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values if =, = ; (, ) (,) (, ) if =, =. So the critical points are,,, and. Now D(, )=D(, )=64 6>0 and f >0 alwas, so f (, )= f (, )= are local minima. 5. f (,)=sin f =sin, f =cos, f =0, f = sin and f =cos. Then f =0 if and onl if =n, n an integer, and substituting into f =0 requires =0 for each of these values. Thus the critical points are ( 0,n ), n an integer. But D(0,n )= cos (n )<0 so each critical point is a saddle point. 6. f (,)=( ) ( ) f =, f =, f =, and f =. Then f =0 implies = or =0 or = and when =, f =0 f = implies =, when =0, f =0 implies =0 or = and when =, f =0 implies =0 or =. Thus the critical points are (,), ( 0,0), (,0), 6

7 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values ( 0,) and (,). Now D(0,0)=D(,0)=D(0,)=D(,)= 6 so these critical points are saddle points, and D(,)=4 with f (,)=, so f (,)= is a local maimum. e f = + e +e =e ( ) 7. f (,)= +, f = + e ()+e =e ( + + ) e ( ( ) ), f =e + =e ( +e ) f =e +()e = 4e ( + ) e. f =e =e ( + + ) ( + )+ ( +e ),, f =0 implies =0, and substituting into f =0 gives e ( )=0 =0 or =. Thus the critical points are ( 0,0) and (,0). D(0,0)=()() 0>0 and f (0,0)=>0, so f (0,0)=0 is a ( 4e ) 0<0 (,0) local minimum. D(,0)= 4e so are saddle points. e f (,)= e f = e +e 8. =, f = e + e = e f = e f = ( ) e f = e,.,, 7

8 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values f =0 implies =0, =0, or =. If =0 then f =0 for an value, so all points of the form 0, are critical points. If =0 then f =0 e =0 =0, so 0,0 (alread included above) is a critical e =0 = = point. If then, so, and, are critical points. D(0,)=0, so the Second Derivatives Test gives no information. However, if >0 then e 0 with equalit onl when =0, so we have local minimum values f (0,)=0, >0. <0 e Similarl, if then 0 with equalit when =0 so f (0,)=0, <0 are local maimum values, and ( 0,0) is a saddle point. D, =8e >0, f, = e / <0 and D, =8e >0 f, = e / >0 f, = /,, so e are local / maimum points while f, = e are local minimum points. 9. f (,)= + + 8

9 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values From the graphs, it appears that f has a local maimum f (0,0) and a local minimum f (0,). There appear to be saddle points near (,). f =6 6, f = + 6. Then f =0 implies =0 or = and when =0, f =0 implies =0 or = ; when =, f =0 implies = or =. Thus the critical points are (0,0), (0,), (,). Now f =6 6, f =6 6 and f =6, so D(0,0)=D(0,)=6>0 while D(,)= 6<0 and f (0,0)= 6, f (0,)=6. Hence (,) are saddle points while f (0,0)= is a local maimum and f (0,)= is a local minimum. f (,)=e 0. There appear to be local maima of about f ( 0.7, 0.7) 0.8 and local minima of about f ( 0.7, 0.7) 0.8. Also, there seems to be a saddle point at the origin. 9

10 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values f =e f =e f =e f =e,,,, f = e ( ) f =0 =0 =. Then implies or. f =0 0,0, Substituting these values into gives the critical points,,,. Then D(,)=e ( ) 4 ( ) ( ) ( ) ( ), so D(0,0)=, while,, D >0 D >0 f, and. But <0, f, f, >0 f,, and <0. Hence ( 0,0) is a saddle point; f, = f, = are local minima and e f, = f, = are local maima. e. f (,)=sin +sin +sin (+), 0, 0 >0 0

11 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values.6 From the graphs it appears that f has a local maimum at about, with value approimatel, a local minimum at about ( 5,5) with value approimatel.6, and a saddle point at about (,). f =cos +cos (+), f =cos +cos (+), f = sin sin (+), f = sin sin (+), f = sin (+). Setting f =0 and f =0 and subtracting gives cos cos =0 or cos =cos. Thus = or =. If =, f =0 becomes cos +cos =0 or cos +cos =0, a quadratic in cos. Thus 5 cos = or and =,, or, ielding the critical points (, ),, and 5, 5. Similarl if =, f =0 becomes cos +=0 and the resulting critical point is (, ). Now D(,)=sin sin +sin sin (+)+sin sin (+). So D(, )=0 and the Second Derivatives Test doesn t appl. D, = 9 4 >0 and f, <0 so f, = is D 5, 5 = 9 4 >0 f 5, 5 5 >0 f, 5 a local maimum while and, so = is a local minimum.. f (,)=sin +sin +cos (+), 0, From the graphs, it seems that f has a local maimum at about 0.5,0.5.

12 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values f =cos sin (+), f =cos sin (+), f = sin cos (+), f = sin cos (+), f = cos (+). Setting f =0 and f =0 and subtracting gives cos =cos. Thus =. Substituting = into f =0 gives cos sin =0 or cos ( sin )=0. But cos 0 for 0 and sin =0 4 implies =, so the onl critical point is 6 6,. Here f 6 6, = <0 and 6 D 6, = 6 4 >0 f. Thus 6, = is a local maimum. 6. f (,)= f (,)=4 0+ and f (,)=. f =0 =0, and the graph of f shows that the roots of f =0 are approimatel =.74, 0. and.40. (Alternativel, we could have used a calculator or a CAS to find these roots.) So to three decimal places, the critical points are (.74,0), (.40,0), and ( 0.,0). Now since f = 0, f =0, f =, and D=4 0, we have D(.74,0)>0, f (.74,0)>0, D(.40,0)>0, f (.40,0)>0, and D(0.,0)<0. Therefore f (.74,0) 9.00 and f (.40,0) 0.4 are local minima, and (0.,0) is a saddle point. The lowest point on the graph is approimatel (.74,0, 9.00). 4. f (,)= f (,)= 0 8, f (,)= Now f =0 = 4, so using a graph, we find solutions to 5 0= f 4, = = (Alternativel, we could have found the roots of f = f =0 directl, using a calculator or a CAS.) To three decimal places, the solutions are.877, 0.45

13 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values ( 0.06, 0.45) and.6, so f has critical points at approimatel.47,.877,, and.04,.6. Now since f = 8, f = 0, f =, and D=96 00, we have D(.47,.877)>0, D(0.06, 0.45)<0, and D(.04,.6)>0. Therefore, since f <0 everwhere, f (.47,.877) 0.8 and f (.04,.6) are local maima, and 0.06, 0.45 is a saddle point. The highest point on the graph is approimatel.47,.877, f (,)= f (,)=+8+ 4, f (,)= Now f =0 ( +)=0 =0 or =. The first of these implies that f = 4 +8+, and the second implies that f =+8+ 4 = From the graphs, we see that the first possibilit for f has roots at approimatel.67, 0.59, and.56, and the second has a root at approimatel.69 (the negative roots do not give critical points, since = must be positive). So to three decimal places, f has critical points at.67,0,,, and. Now since ( 0.59,0) (.56,0) (.69,.06) f =8, f =4, f =4, and D=(8 )(4 ) 6, we have D(.67,0)>0, f (.67,0)>0, D( 0.59,0)<0, D(.56,0)<0, D(.69,.06)>0, and f (.69,.06)<0. Therefore, to three decimal places, f (.67,0).0 and f (.69,.06) 8.05 are local maima, and ( 0.59,0) and (.56,0) are saddle points. The highest points on the graph are approimatel.69,.06,8.05.

14 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values 6. f (,)=e cos f (,)=e and f (,)=4 4sin. From the graphs, we see that to three decimal places, f =0 when 0.459, 0.90, or.7, and f =0 when 0 or (Alternativel, we could have used a calculator or a CAS to find the roots of f =0 and f =0.) So, to ( 0.459, 0.99) ( 0.90,0) three decimal places, f has critical points at 0.459,0,,, ( 0.90, 0.99), (.7,0), and (.7, 0.99). Now f =e 6, f =0, f = 4cos, and ( 4cos ). Therefore D( 0.459,0)<0, D( 0.459, 0.99)>0, f ( 0.459, 0.99)>0 D= e 6, D(0.90,0)>0, f (0.90,0)<0, D(0.90, 0.99)<0, D(.7,0)<0, D(.7, 0.99)>0, and f (.7, 0.99)>0. So f ( 0.459, 0.99).868 and f (.7, 0.99) are local minima, ( 0.90, 0.99) (.7,0) f (0.90,0) 5.7 is a local maimum, and 0.459,0,, and are saddle points. The lowest points on the graph are approimatel.7, 0.99,

15 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values 7. Since f is a polnomial it is continuous on D, so an absolute maimum and minimum eist. Here f =4, f = 5 so there are no critical points inside D. Thus the absolute etrema must both occur on the boundar. Along L, =0 and f (0,)= 5 for 0, a decreasing function in, so the maimum value is f ( 0,0)= and the minimum value is f (0,)= 4. Along L, =0 and f (,0)=+4 for 0, an increasing function in, so the minimum value is f (0,0)= and the maimum value is f (,0)=9. Along L, = + f, and + = 4 for 0, an increasing function in, so the minimum value is f (0,)= 4 and the maimum value is f (,0)=9. Thus the absolue maimum of f on D is f (,0)=9 and the absolute minimum is f (0,)= Since f is a polnomial it is continuous on D, so an absolute maimum and minimum eist. f =, f =, and setting f = f =0 gives, as the onl critical point, where. Along f (,)= L : = and f (,)= for 0 4, a decreasing function in, so the maimum value is f (,0)= and the minimum value is f (,4)=. Along L : =0 and f (,0)= for 5, a decreasing function in, so the maimum value is f (,0)= and the minimum value is 5

16 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values f (5,0)=. Along L, =5 and f (,5 )= +6 7= + for 5, which has a maimum at = where f (,)= and a minimum at both = and =5, where f (,4)= f (5,0)=. Thus the absolute maimum of f on D is f (,0)= f (,)= and the absolute minimum is f (,4)= f (5,0)=. 9. f (,)=+, f (,)=+, and setting f = f =0 gives (0,0) as the onl critical point in D, with f (0,0)=4. On L : =, f (, )=5, a constant. On L : =, f (,)= ++5, a quadratic in which attains its maimum at (,), f (,)=7 and its minimum at,, f, = 9. 4 On L : f (,)= +5 which attains its maimum at (,) and (,) with f (,)=7 and its minimum at (0,), f (0,)=5. On L : f (,)= ++5 with maimum at (,), f (,)=7 and 4 minimum at,, f, = 9. Thus the absolute maimum is attained at both (,) 4 with f (,)=7 and the absolute minimum on D is attained at (0,0) with f (0,0)=4. D 0. f (,)=4 and f (,)=6, so the onl critical point is, (which is in ) where f (,)=. Along L : =0, so f (,0)=4 = +4, 0 4, which has a maimum value when = where f (,0)=4 and a minimum value both when =0 and =4, where f (0,0)= f (4,0)=0. 6

17 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values Along L : =4, so f (4,)=6 = +9, 0 5, which has a maimum value when = where f (4,)=9 and a minimum value when =0 where f (4,0)=0. Along L : =5, so f (,5)= +4+5= +9, 0 4, which has a maimum value when = where f (,5)=9 and a minimum value both when =0 and =4, where f (0,5)= f (4,5)=5. Along L : =0, so f 0, 4 =6 = +9, 0 5, which has a maimum value when = where f (0,)=9 and a minimum value when =0 where f (0,0)=0. Thus the absolute maimum is f (,)= and the absolute minimum is attained at both 0,0 and, where. ( 4,0) f (0,0)= f (4,0)=0. f (,)= is a polnomial and hence continuous on D, so it has an absolute maimum and minimum on D. In Eercise 7, we found the critical points of f ; onl, with is f (,)=0 inside D. On L : =0, f (,0)= 4 +, 0, a polnomial in which attains its maimum at =, f (,0)=8, and its minimum at =0, f (0,0)=. On L : =, f (,)= 4 +8, 0, a polnomial in which attains its minimum at =, f (, )= , and its maimum at =0, f (,0)=8. On L : =, f (,)= 4 8+8, 0, a polnomial in which attains its minimum = at, f (, )= , and its maimum at =, f (,)=75. On L : =0, 4 f (0,)= 4 +, 0, a polnomial in which attains its maimum at =, f (0,)=8, and its minimum at =0, f (0,0)=. Thus the absolute maimum of f on D is f (,0)=8 and the absolute 7

18 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values minimum is f (,)=0.. f = and f =, and since f =0 =0, there are no critical points in the interior of D. Along L, =0 and f (,0)=0. Along L,=0 and f (0,)=0. Along L, =, so let g()= f (, )= for 0. Then g / = =0 =. The maimum value is f (, )= and the minimum occurs both at =0 and = where f ( 0, )= f (,0)=0. Thus the absolute maimum of f on D is f (, )=, and the absolute minimum is 0 which occurs at all points along L and L.. f (,)=6 and f (,)=4. And so f =0 and f =0 onl occur when ==0. Hence, the onl critical point inside the disk is at ==0 where f (0,0)=0. Now on the circle + =, = so let g()= f (,)= + ( ) = g / ()=4 +6,. Then 4=0 =0,, or. f (0, )=g ( 0)= f,, =g =, and (, ) is not in D. Checking the 6 endpoints, we get f (,0)=g= and f (,0)=g()=. Thus the absolute maimum and minimum of f on D are f (,0)= and f (,0)=. Another method: On the boundar + = we can write =cos, =sin, so f (cos,sin )=cos +sin 4, 0. (, ) (,) (, ) 4. f (,)= and f (,)= + and the critical points are,,,, and. But onl (,) and (,) are in D and f (,)=4, f (,)=8. Along L : = and f (,)= +,, which has a maimum at = where f (,)=4 and a minimum at = where f (, )= 8. Along L : = and f (,)= +,, which has a maimum at = where f (,)=8 and a minimum at = where f (,)=. Along L : = and f (,)= +9,, which has a maimum at = and = where f (,)= f (,)= and a minimum at = 8

19 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values and = where f (,)= f (,)=7. Along L : = and f (,)=9, 4, which has a maimum at = where f (,)=8 and a minimum at = where f (, )= 8. So the absolute maimum value of f on D is f (,)=8 and the minimum is f (, )= f (,)= ( ) ( ) f (,)= () ( ) and f (,)= f (,)=0 =0 =0. Setting gives either or. There are no critical points for =0, since f (0,)=, so we set =0 = + ( 0 ), so f, + = ( )() + + = 4 ( ). Therefore f (,)= f (,)=0 at the points, and. To classif these critical points, we calculate (,0) f (,)= ++4+, f (,)= 4, and f (,)= In order to use the Second Derivatives Test we calculate D(,0) = f (,0) f (,0) [ f (,0)] =6>0, (,) f (,0)= 0<0, D(,)=6>0, and f (,)= 6<0, so both,0 and give local maima. 6. f (,)=e e is differentiable everwhere, so the requirement for critical points is that () f =e =0 and () f =e e =0. From () we obtain e =, and then () gives 6 =0 9

20 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values = or 0, but onl = is valid, since =0 makes () impossible. So substituting = into () gives =0, and the onl critical point is,0. The Second Derivatives Test shows that this gives a local maimum, since ( e ),0 D(,0)= 6 e 9e =7>0 and f (,0)= 6 = 6<0. But f (,0)= is not an (,0) absolute maimum because, for instance, f (,0)=7. This can also be seen from the graph. (,,z) + z= 7. Let d be the distance from,, to an point on the plane, so d= + +(z+) where z=+, and we minimize d = f (,)= + +(+). Then f (,)=+(+)=4+ 4, f (,)=+(+)=+4. Solving 4+ 4=0 and +4 =0 simultaneousl gives =, =0. An absolute minimum eists (since there is a minimum distance from the point to the plane) and it must occur at a critical point, so the shortest distance occurs for =, =0 for which d= +(0 ) +(+0) =. d= + +(z ) 8. Here the distance d from a point on the plane to the point,, is, where z=4 +. We can minimize d = f (,)= + +( +), so f (,)=+( +)=4 4 and f (,)= + ( +)=4. Solving 4 4=0 4 =0 = 5 = 4 5 and simultaneousl gives and, so the onl critical point is, 4. This point must correspond to the minimum distance, so the point on the plane closest to (,,) is 5, 4,. 9. Minimize d = + +z = Then f =+, f =+ so the critical point is 0,0 and D(0,0)=4 >0 with f (0,0)= so this is a minimum. Thus z = or z= and the points on the surface are ( 0,0, ). 0

21 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values d = + +z = = f (,) f = z=/ Since on the surface, we minimize. 5 4, 4 f = 4 5, so the critical points occur when = and = or 6 4 == 4 6, so = = and 0 = = /0, = /0. The four critical points are ( /0, /0 ). The absolute minimum must occur at these points (there is no maimum since the surface is infinite in etent). Thus the points on the surface closest to the origin are ( /0, /0, /5 ) z=00, so maimize f (,)=(00 ). f =00, f =00, f =, f =, f =00. Then f =0 implies =0 or =00. Substituting =0 into f =0 gives =0 =00 =00 f =0 00 or and substituting into gives 00=0 so =0 or. Thus the 00 critical points are ( 0,0), ( 00,0), ( 0,00) and, 00. D(0,0)=D(00,0)=D(0,00)= 0, 000 while D 00, 00 = 0,000 and 00 f, = <0. Thus 0,0, ( 00,0) and ( 0,00) are saddle points whereas 00 f, 00 is a local maimum. Thus the numbers are ==z= 00. c 4. Maimize f (,)= a b 00. f =a a b (00 ) c c a b (00 ) c = a b (00 ) c [a(00 ) c] and f = a b (00 ) c [b(00 ) c]. Since, and z are all positive, the onl critical point =a 00 = 00b 00a a+c a+b+c a+b+c, 00b occurs when and. Thus the point is and the numbers are a+b+c = 00a, = 00b, z= 00c. a+b+c a+b+c a+b+c / / (,,z) / ( = ) ( ) / 4. Maimize f (,)= with in first octant. Then / f = and

22 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values f = / f =0 =0 =. Setting gives or but >0, so onl the latter solution applies. Substituting this into f =0 gives = 4 or =, = and then z =(6 )/4=. The fact that this gives a maimum volume follows from the geometr. This maimum volume is V =()()(z)=8 ( )= 6. / f (,)= a b c b c a c 44. Here maimize a b. Then and / f =c a. Then f =0 a b ( a b c b c a c ) / f =c a b b a a b a b c b c a c b a b >0 = a b b (with, ) implies a and substituting into f =0 implies b =a b or = a, = b and then z= c. Thus the maimum volume of such a rectangle is V =()()(z)= 8 abc. 45. Maimize f (,)= (6 ), then the maimum volume is V =z. f = 6 = (6 ) and f = 6 4. Setting f =0 and f =0 gives the critical point (,) which geometricall must ield a maimum. Thus the volume of the largest such bo is V =()() = Surface area =(+z+z)=64 cm, so +z+z= or z=. Maimize the volume + f (,)= f =. Then + (+) = (+) and f = (+). Setting f =0 ( ) ( 4 ) ( ) =0 implies = and substituting into f =0 gives 4 or () =0. Thus = 64 or = 8, = 64/ 6 6 6/ 6 = 8 and z= 8. Thus the bo is a cube 6 6

23 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values with edge length 8 6 cm. 47. Let the dimensions be,, and z ; then 4+4+4z=c and the volume is V =z= 4 c = 4 c, >0, >0. Then V = 4 c and V = 4 c, so V =0=V when += 4 c and += 4 c. Solving, we get = c, = c and z= 4 c = c. From the geometrical nature of the problem, this critical point must give an absolute maimum. Thus the bo is a cube with edge length c. 48. The cost equals 5+(z+z) and z=v, so C(,)=5+V (+)/()=5+V +. Then f =0 = 5 V = C =5 V, C =5 V, f =0 implies =V / 5, implies. Thus the dimensions of the aquarium which minimize the cost are == 5 V units, z=v / 5. / 49. Let the dimensions be, and z, then minimize +(z+z) if z=, 000 m. Then f (,)=+[64], 000(+)/]=+64, 000( + ), f = 64, 000, f = 64, 000. And f =0 implies =64, 000/ ; substituting into f =0 implies =64, 000 or =40 and then =40. Now f (40,40)>0 D(,)=[()(64], 000)] >0 for 40,40 and so this is indeed a minimum. Thus the dimensions of the bo are ==40 cm, z=0 cm. 50. Let be the length of the north and south walls, the length of the east and west walls, and z the height of the building. The heat loss is given b h=0(z)+8(z)+()+5()=6+6z+0z. The volume is 4000 m, so z=4000, and we substitute z= 4000 to obtain the heat loss function h(,)=6+80, 000/+64, 000/. (a) Since z= , /. Also 0 and 0, so the domain of h is D={(,) 0,0 000/}.

24 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values (b) h(,)=6+80, , 000 h =6 80, 000, h =6 64, 000. h =0 implies 6 =80, 000 = 80,000 6 and substituting into h =0 gives 6=64, ,000 = 80, ,000 = 50,000 = 50,000 = , so =, 60 h 0 50, 80 and the onl critical point of is ( 5.54,0.4) which is not in D. Net we check the boundar of D. On L : =0, h(,0)=80+80, 000/+6400/, 0. Since h / (,0)= / 00, >0 for 0, h(,0) is an increasing function with minimum 00 h(0,0)=0, 00 and maimum h,0 0, 5. On L : =000/, 00 h(,000/)= , 000/, 0. Since h / (,000/)=64 80, 000/ <0 for , h(,000/) is a decreasing function with minimum h,0 0, 5 and h 0, maimum 0, 587. On L : =0, h(0,)=80+64, 000/+8000/, 0. h / (0,)= / 00, >0 for 0, so h(0,) is an increasing function of with minimum h(0,0)=0, 00 and maimum h 0, 00 0, 587. Thus the absolute minimum of h is h(0,0)=0, 00, and the dimensions of the building that minimize heat loss are walls 0 m in 4000 length and height 0 = m. (c) From part (b), the onl critical point of h, which gives a local (and absolute) minimum, is approimatel h(5.54,0.4) 996. So a building of volume 4000 m with dimensions 5.54 m, 0.4 m, 4

25 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values z (5.54)(0.4) m has the least amount of heat loss. 5. Let,, z be the dimensions of the rectangular bo. Then the volume of the bo is z and L= + +z L = + +z z= L. Substituting, we have volume V (,)= L,,>0. / + L = L V = L, L V = L, L V =0 implies L = ( L )=0 + =L (since >0 ), and V =0 implies ( L )= ( L )=0 + =L (since >0 ). Substituting =L into =L/ + =L gives +L 4 =L =L =L/ (since >0 ) and then = L L/. So the onl critical point is ( L/,L/ ) which, from the geometrical nature of the problem, must give an absolute maimum. Thus the maimum volume is V ( L/,L/ )=( L/ ) L ( L/ ) ( L/ ) =L / ( ) 5. Since p+q+r= we can substitute p= r q into P giving P=P(q,r)=( r q)q+( r q)r+rq=q q +r r rq. Since p, q and r represent proportions and p+q+r=, we know q 0, r 0, and q+r. Thus, we want to find the absolute maimum of the continuous function P(q,r) on the closed set D enclosed b the lines q=0, r=0, and q+r=. To find an critical points, we set the partial derivatives equal to zero: P (q,r)= 4q r=0 and q P (q,r)= 4r q=0. The first equation gives r= q, and substituting into the second equation we r have 4( q) q=0 q=,. Then we have one critical point,, where P, =. Net we find the maimum values of P on the boundar of D which consists of three line segments. For the segment given b r=0, 0 q, P(q,r)=P(q,0)=q q, 0 q. This represents a parabola with maimum value P,0 =. On the segment q=0, 0 r we have P(0,r)=r r, 0 r. This represents a parabola with maimum value P 0, =. Finall, on the segment q+r=, 0 q, cubic units. 5

26 Stewart Calculus ET 5e 05497;4. Partial Derivatives; 4.7 Maimum and Minimum Values P(q,r)=P(q, q)=q q, 0 q which has a maimum value of P, =. Comparing these values with the value of P at the critical point, we see that the absolute maimum value of P(q,r) on D is. =0 n =0 n implies n 5. Note that here the variables are m and b, and f (m,b)= i = i m i +b. Then n f m = i = m +b i i i i = i i m i b n =0 i i = i =m n i i = +b n implies or i i = and i f b = n i = i m i +b f mm = n i = i =m n i = i + n i = b=m n f bb = n i = i +nb. Thus we have the two desired equations. Now i =, i i = =n and i =. And f (m,b)>0 i mm n n n n alwas and D(m,b)=4n 4 =4 n >0 alwas so the i = i i = i n i = i solutions of these two equations do indeed minimize. i = d i f mb = n 54. An such plane must cut out a tetrahedron in the first octant. We need to minimize the volume of the tetrahedron that passes through the point (,,). Writing the equation of the plane as a + b + z c =, the volume of the tetrahedron is given b V = abc. But (,,) must lie on the plane, 6 so we need a + b + c = ( * ) and thus can think of c as a function of a and b. Then V = b c+a c and V = a c+b c. Differentiating ( * ) with respect to a we get a 6 a b 6 b a c c a =0 i = i c c = a a * b b c, and differentiating with respect to gives c b =0 c c = b b V = b c c+a =0 c=a a 6 a V = a c. Then, and c+b =0 b 6 b c= b. Thus a= b b=a * or. Putting these into gives = or a= and then b=6, c=9. Thus the equation of a the required plane is z = or 6++z=8. 9 6