MA261-A Calculus III 2006 Fall Homework 8 Solutions Due 10/30/2006 8:00AM

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1 MA61-A Calculus III 006 Fall Homework Solutions Due 10/0/006 :00AM 116 # Let f (x; y) = y ln x (a) Find the gradient of f (b) Evaluate the gradient at the oint P (1; ) (c) Find the rate of change of f at P in the direction of the vector u = 4 ; (a) The gradient is rf (x; y) ; D y E = x ; ln x (b) The gradient at the oint P (1; ) is rf (1; ) = 1 ; ln 1 = h ; 0i (c) The directional derivative is D u f (x; y) = rf (x; y) u At the oint P (1; ), we have 4 D u f (1; ) = rf (1; ) ; = h ; 0i 1 4 ; = 1

2 116 #1 Find the directional derivative of the function f (x; y) = ln x + y at the oint (; 1) in the direction of the vector v = h 1; i First, we need an unit vector of v The unit vector u = v Second, the gradient of f is rf (x; y) ; = x x + y ; Thus, the directional derivative at the oint (; 1) is 1 4 D u f (1; ) = rf (; 1) ; = ; 116 #1 Find the directional derivative of the function f (x; y; z) = x + y + z D = ( 1) + jvj = h 1;i y x + y 1 ; = 0 1 ; E at the oint P (; 1; ) in the direction of the origin First, the directional vector is v = h0; 0; 0i h; 1; i = h ; 1; i But, we need an unit vector of v The unit vector u = v Second, the gradient of f is rf (x; y; z) = jvj = D h ; 1; i = () +(1) ; ; = hx; y; zi Thus, the directional derivative at the oint (; 1; ) is 1 D u f (; 1; ) = rf (; 1; ) ; ; = h4; ; 6i #0 Find the maximum rate of change of f (; q) = qe + e q at the oint (0; 0) in the direction in which it occurs The gradient is rf (; q) = qe + e q ; e e q 14 ; 1 14 ; which is the direction with the maximum rate of change At (0; 0), the rate is q jrf (0; 0)j = jh1; 1ij = (1) + (1) = 1 14 ; 14 ; 14 E 14 = 14

3 116 #4 Find the directions in which the direction derivative of f (x; y) = x + sin xy at the oint (1; 0) has the value 1 Assume that the direction is ha; bi where 1 = jha; bij = a + b We know that the gradient rf = hx + y cos xy; x cos xyi The directional derivative at the oint (1; 0) is D u f (1; 0) = rf (1; 0) ha; bi = h; i ha; bi = a + b If the directional derivative has the value 1, we have a + b = 1, that is, b = 1 a Note that we need a unit vector to calculate the directional derivative So, we need a + 1 a = 1 This tells us that a 4a = 0, or, a = 1 7 D So, we have two 4 1+ directions 7; 1 E D 7 1 or E 7; #0 Suose you are climbing a hill whose shae is given by the equation z = :00x 0:01y, where x, y, z are measured in meters, you are sting at a oint which coordinates (60; 40; 966) The ositive x-axis oint oints east the ositive y-axis oints north (a) If you walk due south, will you start to ascend or descend? At what rate? (b) If you walk northwest, will you start to ascend or descend? At what rate? (c) In which direction is the sloe largest? What is the rate of ascent in that direction? At what angle above the horizontal does the ath in that direction begin? Consider f (x; y) = :00x 0:01y We have f x = 0:01x f y = 0:0y (a) We can use a unit vector h0; 1i to reresent the direction toward south So, the directional vector is D u f (60; 40) = rf (60; 40) h0; 1i = h 0:6; 0:i h0; 1i = 0: So, you start to ascend at the rate 0: (b) We can use D a vector h 1; 1i to reresent the direction toward northwest The unit 1 vector is 1 ; E So, the directional vector is D u f (60; 40) = rf (60; 40) ; = h 0:6; 0:i ; = 0: So, you start to descend at the rate 0: (c) The gradient is rf (60; 40) = h 0:6; 0:i q ( 0:6) + ( 0:) = which is the direction with the largest sloe The rate is jh 0:6; 0:ij = 1 Since the rate is 1 means that when you move 1 meter, you climb u 1 meter So, the angle here is tan = 1 This tells us that = #6 Find the equation of the tangent lane the normal line to the surface at the oint (1 + ; 1; 1) x z = 4 arctan (yz)

4 4 Let F (x; y; z) = x z 4 arctan (yz) Then the surface becomes a level surface F = 0 To write these two equations, we need the gradient of F (x; y; z) which rf (x; y; z) 4z = 1; 1 + (yz) ; 1 4y 1 + (yz) [Note that d 1 (arctan t) = ] So, at the oint (1 + ; 1; 1), the gradient is dt 1+t 4 1 rf (1 + ; 1; 1) = 1; 1 + (1 1) ; (1 1) = h1; ; i or, Thus, the tangent lane equation is (1) (x (1 + )) + ( ) (y 1) + ( ) (z 1) = 0, The normal line equation is x y z + 4 = 4 x (1 + ) = y 1 1 = z #44 Find the oints on the ellisoid x + y + z = 1 where the tangent lane is arallel to the lane x y + z = 1 We can treat the ellisoid x + y + z = 1 as a level surface F (x; y; z) = 1 where F (x; y; z) = x + y + z According to the tangent lane equation in textbook age 796, the normal vector of the tangent lane to the level surface F (x; y; z) = 1 is rf = hx; 4y; 6zi If the tangent lane is arallel to the lane x y + z = 1, their normal vector must be arallel Thus, hx; 4y; 6zi = k h; 1; i This imlies that hx; y; zi = k; 1 k; k Since this oint must be on the ellisoid, we have 4 6 k k + 6 k = 1 By solving it, we have k =, or, k = Therefore, there are two oints! ; 10 ;! ; 10 ; 116 #46 Show that the ellisoid x +y +z = 9 the shere x +y +z x 6y z+4 = 0 are tangent to each other at the oint (1; 1; ) (This means that they have a common tangent lane at the oint) We can treat the ellisoid x + y + z = 9 as a level surface F (x; y; z) = 9 where F (x; y; z) = x + y + z According to the tangent lane equation in textbook age 796, the normal vector of the tangent lane to the level surface F (x; y; z) = 9 is rf = h6x; 4y; zi Similarly, we can treat the shere x + y + z x 6y z + 4 = 0 as a level surface G (x; y; z) = 0 where G (x; y; z) = x + y + z x 6y z + 4 According to the

5 tangent lane equation in textbook age 796, the normal vector of the tangent lane to the level surface G (x; y; z) = 0 is rg = hx ; y 6; z i At the oint (1; 1; ), rf (1; 1; ) = h6; 4; 4i rg (1; 1; ) = h 6; 4; 4i Since these two normal vectors generate the same tangent lane at (1; 1; ), we know that our two grah share a common tangent lane at (1; 1; ) Thus, the ellisoid x +y +z = 9 the shere x + y + z x 6y z + 4 = 0 are tangent to each other at the oint (1; 1; ) 116 #4 Show that every normal line to the shere x + y + z = r asses through the center of the shere We can treat the shere x + y + z = r as a level surface F (x; y; z) = r where F (x; y; z) = x + y + z According to the normal line equation in textbook age 796, the directinoal vector of the normal line to the level surface F (x; y; z) = r is rf = hx; y; zi For a oint (x 0 ; y 0 ; z 0 ) on the shere, the normal line equation at that oint is x x 0 = y y 0 = z z 0 x 0 y 0 z 0 We notice that the center of the shere, which is the origin (0; 0; 0), satis es this normal line equation Therefore, this normal line asses through the center of the shere Since (x 0 ; y 0 ; z 0 ) is chosen arbitrarily, we rove that every normal line to the shere x +y +z = r asses through the center of the shere 117 #6 Find the local maximum minimum values saddle oint(s) of the function First, we have rf (x; y) = f (x; y) = x y + 1x ; = x y + 4x; x By assuming rf (x; y) = 0, we have x y + 4x = 0 x = 0 x = 0 imlies that x = Thus, we have () y + 4 () = 0, or, y = 4 So, there is only one critical oint (; 4) To classify it, we need the determint = f xx f yy f xy = (6xy) (0) x = 9x 4 Since < 0 at the critical oint (; 4), we know that it is a saddle oint 117 #14 Find the local maximum minimum values saddle oint(s) of the function f (x; y) = x x y y

6 6 First, we have rf (x; y) ; = ( x) y y ; x x ( y) By assuming rf (x; y) = 0, we have ( x) (y y ) = 0 (x x ) ( y) = 0 From the rst equation, we have x = 0, or y y = 0 This tells us that x = 1, or, y = 0 or Similarly, From the second equation, we have x x = 0, or y = 0 This tells us that x = 0 or, or, y = 1 There are several combinations: (a) If x = 1, then y must be 1 from the second equation (b) If y = 0, then x can be 0 or from the second equation (c) If y =, then x can be 0 or from the second equation Therefore, we have severl critical oints (1; 1), (0; 0), (; 0), (0; ) (; ) To classify them, we need the determint (x; y) = f xx f yy f xy = ( ) y y x x ( ) (( x) ( y)) = 4xy ( y) ( x) (1 x) (1 y) f xx (x; y) = ( ) (y y ) = y ( y) (a) For the critical oint (1; 1), (1; 1) = 4 > 0 f xx (1; 1) = < 0 Thus, it is a local maximum (b) For the critical oint (0; 0), (0; 0) = < 0 Thus, it is a saddle oint (c) For the critical oint (; 0), (; 0) = < 0 Thus, it is a saddle oint (d) For the critical oint (0; ), (0; ) = < 0 Thus, it is a saddle oint (e) For the critical oint (; ), (; ) = < 0 Thus, it is a saddle oint The function looks like 10 4 y 1 z x #6 Find the absolute maximum minimum values of f (x; y) = + xy x y on the set D whish is the closed triangular region with vertices (1; 0), (; 0), (1; 4)

7 D can be seen as a closed region bounded by x = 1, y = 0 x + y = (the line asses through (; 0), (1; 4)) For the interior art of D, we calculate the rf = hy 1; x i rst If rf = 0, then we have (x; y) = (; 1) It is a critical oint with value f (; 1) = 1 For the boundary art, there are three di erent edges, E 1 = f(x; 0) j 1 x g, E = f(1; y) j 0 y 4g, E = f(x; x) j 1 x g For E 1, our function becomes f (x; 0) = x Thus, with 1 x, we have f (x; 0) The maximum haens at (1; 0) the minimum haens at (; 0) For E, our function becomes f (1; y) = y Thus, with 0 y 4, we have f (1; y) The maximum haens at (1; 0) the minimum haens at (1; 4) For E, our function becomes f (x; x) = +x ( x) x ( x) = x +6x 7 If the derivative x+6 is zero, we have x = So, there are three cidates for extrema: x = 1; ; When x = 1, y = 4 f (1; 4) = When x =, y = f (; ) = When x =, y = 0 f (; 0) = Thus, we have the maximum haens at (; ) the minimum haens at (1; 4) (; 0) Hence, the absolute maximum is which is haened at (1; 0) (; ) the absolute minimum is which is haened at (; 0) (1; 4) 117 #0 Find the absolute maximum minimum values of f (x; y) = xy on the set D = f(x; y) j x 0; y 0; x + y g For the interior art f(x; y) j x > 0; y > 0; x + y < g, we calculate the rf = hy ; xyi rst If rf = 0, then we have (x; y) = (0; 0) It sits on the boundary, not in our interior art For the boundary art, there are three di erent edges The rst one is B 1 = f(x; 0) j 0 < x < g In B 1, we have 0 f (x; y) 0 Thus, f (x; y) = 0 The second one is B = f(0; y) j 0 < y < g In B, we have 0 f (x; y) 0 Thus, f (x; y) = 0 The third one is B = f(x; y) j x > 0; y > 0; x + y = g On B, our function becomes f (x; y) = xy = x ( x ) Let g (x) = x ( x ) = x x To nd the extrema, we set 0 = g 0 (x) = x This imlies that x = 1, or, x = 1 (since x > 0) When x = 1, on the B, we have y = (since y > 0) Therefore, the extreme haens at 1; with the value f 1; = So, the maximum is at 1; the minimum is 0 when x = 0 or y = 0 [Another way to get the extreme on B ] We can use the olar coordinate to reresent B This gives us that B = (r; ) j r = 0 < < where x = cos y = sin Thus, we have f = xy = cos sin = cos sin Set g () = cos sin To nd the maximum the minimum values, we consider g 0 () = 0, that is, ( sin ) sin + cos ( sin cos ) = ( sin ) sin + cos Since 0 < <, we know that sin 6= 0 Thus, sin + cos = 0, or sin = cos (since 0 < < ) This imlies that tan = Therefore, we have sin = cos = 1 It means that the extreme haens at 1; with the value f 1; = 7

8 117 #6 Find the oints on the surface y = 9 + xz that are closest to the origin Let (x; y; z) be a oint on the surface y = 9 + xz The distance to the origin is d = q (x 0) + (y 0) + (z 0) = x + y + z To minimize the distance, it is equivalent to consider d = x +y +z Since y = 9+xz, we have d = x + (9 + xz) + z = x + xz + z + 9 Let f (x; z) = x + xz + z + 9 We have f x (x; z) = x + z f z (x; z) = x + z If f x = f y = 0, we have x + z = 0 x + z = 0 These two equations imly that there is only one critical oint (x; z) = (0; 0) Also, D = f xx f zz fxz = 1 = > 0 f xx = > 0 Thus, (x; z) = (0; 0) is a local minimum It gives us y = = 9, or, y = So, there are two oints (0; ; 0) (0; ; 0) on the surface y = 9 + xz that are closest to the origin 117 # Find three ositive numbers x, y, z whose sum is 100 such that x a y b z c is a maximum Assume that all a, b, c are not zero a + b + c 6= 0 Since the sum of x, y, z is 100, we have x + y + z = 100 Assuma that f (x; y; z) = x a y b z c To maximize f under the constraint g (x; y; z) = x + y + z 100 = 0, we use the lagrange multilier Assume that rf = rg This tells us that ax a 1 y b z c ; x a by b 1 z c ; x a y b cz c 1 = h1; 1; 1i So, we have ax a 1 y b z c = x a by b 1 z c = x a y b cz c 1 x + y + z = 100 Obviously, if f (x; y; z) is a maximum, then all x, y, z are not zero If all x, y, z are not one, then we have ayz = bxz = cxy This imlies that x = ay z = c y By utting it back to the g = 0, we have y = 100b b b Thus, x = 100a 100c z = a+b+c a+b+c a+b+c 117 #4 Find the dimensions of the rectangular box with largest volume if the total surface area is given as 64 cm Assume that the dimensions of the box is (x; y; z) where x > 0, y > 0, z > 0 The volume is V = xyz The surface area is xy + yz + xz = 64 It tells us that z = xy x + y

9 xy Consider f (x; y) = xy We have x+y xy ( y) (x + y) ( xy) (1) f x (x; y) = y + xy x + y (x + y) xy y = y + xy x + y (x + y) = y ( x + xy) (x + y) xy ( x) (x + y) ( xy) (1) f y (x; y) = x + xy x + y (x + y) xy x = x + xy x + y (x + y) = x ( y + xy) (x + y) Note that x+y 6= 0 since x y all ositive If f x = 0 = f y, we have x + xy = 0 y + xy = 0 since x 6= 0 y 6= 0 This tells us that x = y, or x = y(again since x y are all ositive) Similarly, if we write x in terms of y z, we will get y = z So, we can conclude q that x = y = z Then, since xy + yz + xz = 64, we have 6x = 64, or x = = y = z [Alternative Solution] Assume that the dimensions of the box is (x; y; z) The surface area is xy + yz + xz = 64 The volume is V = xyz To maximize V under the constraint g (x; y; z) = xy + yz + xz 64 = 0, we use the lagrange multilier Assume that rf = rg This tells us that hyz; xz; xyi = h (y + z) ; (x + z) ; (x + y)i So, we have xyz = (xy + xz) = (xy + yz) = (xz + yz) If 6= 0, then xy + xz = xy + qyz = xz + yz This imlies that x = y = z By the constraint, we have x = y = z = If = 0, then hyz; xz; xyi = 0 Therefore, the volume is 0 which is not what we want 117 #0 Find an equation of the lane that asses through the oint (1; ; ) cut o the smallest volume in the rst octant An equation of the lane that asses through the oint (1; ; ) looks like a (x 1) + b (y ) + c (z ) = 0 We need the intersection oints of the lane three axes to calculate the volume Notice that if any of a; b; c equals zero, this lane contains one of the three axes the volume becomes in nity So, we can assume that a 6= 0, b 6= 0, c 6= 0 By assuming a+b+c y = 0 z = 0, the intersection oints of the lane the x-axis is ; 0; 0 a 9

10 10 Similarly, the intersection oints of the lane the y-axis is 0; a+b+c ; 0 the b intersection oints of the lane the z-axis is 0; 0; a+b+c c Therefore, the volume is V = 1 a + b + c a + b + c a + b + c 6 a b c since it looks like a yramid 11 #4 Use Lagrange multiliers to nd the maximum minimum values of the function subject to the constraint Let g (x; y) = x + y f (x; y) = 4x + 6y x + y = 1 1 First, we calculate rf (x; y) ; = h4; rg (x; y) = hx; yi If rg (x; y) = 0, then (x; y) = (0; 0) which does not satisfy the constraint g = 0 Assume that rf = rg With the constraint, we have < : 4 = x 6 = y x + y = 1 From the rst second equations, we know that x 6= 0, y 6= 0 6= 0 We can solve them to get x = 4 = y = 6 = By lugging into the constraint, we have + = 1 This imlies that = 1, or, = 1 Thus, we have x = y = Therefore, we have two critical oints (; ) ( ; ) The values of the function f are f (; ) = 6 f ( ; ) = 6 Hence, the maximum is 6 at the oint (; ) the minimum is 6 at the oint ( ; ) 11 #1 Use Lagrange multiliers to nd the maximum minimum values of the function f (x; y; z) = x 4 + y 4 + z 4 subject to the constraint x + y + z = 1 Let g (x; y; z) = x + y + z 1 First, we calculate rf (x; y; z) ; ; = 4x ; 4y ; rg (x; y; z) = hx; y; zi If rg (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does not satisfy the constraint g = 0

11 Assume that rf = rg With the constraint, we have 4x >< = x 4y = y 4z >: = z x + y + z = 1 Notice that if = 0, then x = 0, y = 0 z = 0 It is not the case we need to consider since (x; y; z) = (0; 0; 0) which does not satisfy the constraint g = 0 Thus, we know that 6= 0 Assume that x 6= 0, y 6= 0 z 6= 0 Then we have >< >: x = y = z = x + y + z = 1 By lugging into the constraint, we have + + = 1 This imlies that = Thus, we have x = 1, y = 1 z = 1 Therefore, we have critical oints in all the combinations The values of the function in this case are all f 1 ; 1 ; 1 = 1 If x = 0, then we have < 4y = y 4z = z : y + z = 1 Similarly, assume that y 6= 0 z 6= 0 Then we have < y = z : = y + z = 1 By lugging into the constraint, we have + = 1 This imlies that = 1 Thus, we have y = 1 z = 1 Therefore, we have 4 critical oints in all the combinations The values of the function in this case are all f 0; 1 ; 1 = 1 If y = 0, then we have z = 1, or, z = 1 We have critical oints in all the combinations The values of the function in this case are all f (0; 0; 1) = 1 Similarly, if y = 0, we have critical oints 1 ; 0; 1 with f 1 ; 0; 1 = 1 (0; 1; 0) with f (0; 1; 0) = 1 If z = 0, we have critical oints 1 ; 1 ; 0 with f 1 ; 1 ; 0 (0; 0; 1) with f (0; 0; 1) = 1 = 1 By combining everything together, we have the maximum is 1 at the oints (0; 0; 1), (0; 1; 0) (1; 0; 0) the minimum is 1 at the oints in all conbinations of 1 ; 1 ; 1 11 #16 Use Lagrange multiliers to nd the maximum minimum values of the function subject to the constraints f (x; y; z) = x y z x + y z = 0 11

12 1 x + z = 1 Let g (x; y; z) = x + y + z 1 h (x; y; z) = x + z 1 First, we calculate rf (x; y; z) ; ; = h; 1; rg (x; y; z) = h1; 1; rh (x; y; z) = hx; 0; 4zi Note that rg (x; y; z) 6= 0 If rh (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does not satisfy the constraint h = 0 Assume that rf = rg + rh With the constraints, we have = + x >< 1 = = + 4z x + y z = 0 >: x + z = 1 It becomes >< >: x = = 1 z = 1 y = z x x + z = 1 By lugging into the last constraint, we have + 1 = 1 This imlies that = 9, or, = Thus, x =, z = y = z x = Therefore, there 6 are two critical oints ; ; ; ; 6 6 The values of the function f are f ; ; = f ; ; = 6 6 Hence, the maximum is at the oint ; ; the minimum is 6 at the oint ; ; 6 11 #1 Find the extreme values of f (x; y) = x + y 4x on the region described by x + y 16 Let g (x; y) = x + y 16 First, we calculate rf (x; y) ; = h4x 4; 6yi

13 rg (x; = hx; yi First, we nd the extreme values for the interior of x + y 16 Consider the region x + y < 16 Assume that rf = 0 This imlies that (x; y) = (1; 0) Since (1; 0) satis es x + y < 16, (1; 0) is a critical oint The value is f (1; 0) = 7 Next, we consider the boundary x + y = 16 Thus, g = 0 becomes a constraint If rg (x; y) = 0, then (x; y) = (0; 0; 0) which does not satisfy the constraint g = 0 Assume that rf = rg With the constraint, we have < : 4x 4 = x 6y = y x + y = 16 If y 6= 0, then by the second equation, we have = This imlies that x = y = 1 In this case, we have two critical oints ; ; The values are both f ; = f ; = 47 If y = 0, then x = 4 In this case, we have two critical oints (4; 0) ( 4; 0) The values of them are f (4; 0) = 11 f ( 4; 0) = 4 Hence, the maximum is 47 at the oints ; ; the minimum is 7 at the oint (1; 0) 11 # Find the maximum minimum volumes of a rectangular box whose surface area is 100 cm whose total edge length is 00 cm Assume that the box has edges x, y z where x > 0, y > 0 z > 0 Thus, the volume can be written as V (x; y; z) = xyz The surface area is xy + yz + xz = 100 This gives us a constraint xy +yz +xz = 70 Also, the total edge length is 4x+4y +4z = 00 This gives us another constraint x + y + z = 0 Let us rewrite our roblem as the following: to maximize or to minimize F (x; y) = xy (0 x y) = 0xy x y xy subject to the constraint xy + y (0 x y) + x (0 x y) = 70, or, 0x + 0y xy x y = 70 Let G (x; y) = 0x + 0y xy x y 70 First, we rf (x; y) = 0y xy y ; 0x xy x rg (x; = h0 y x; 0 x yi If rg (x; y) = 0, then we have 0 y x = 0 = 0 x y It imlies that x = y = 0 which does not satisfy the constraint G = 0 Assume that rf = rg With the constraints, we have < : 0y xy y = (0 y x) 0x xy x = (0 x y) 0x + 0y xy x y = 70 1

14 14 If 0 y x 6= 0, then = y So, the second equation becomes 0x xy x = y (0 x y), or, (0 y x) (x y) = 0 Thus, we have x = y or 0 y x = 0 If x = y, then the constraint becomes x 100x + 70 = 0, or, x = = y The values are F 10; :979 F :447 If 0 y x = 0, then the constraint becomes 0 (0 y) + 0y (0 y) y (0 y) y = 0, or, y (y 100) = 0 Since y is an edge, y > 0 So, we have = ; y = Therefore, x = which is a contradiction since x is an edge x > 0 Since x; y; z are symmetric in our equations, by assuming y = z or x = z, we can have another two sets of critical oints 10; ; , ; ; , 10; ; Hence, the maximum is :447 with edges the minimum is 947:979 at the oint ; ; ; ; ; ; [Alternative Solution] Let g (x; y; z) = xy + yz + xz 70 h (x; y; z) = x + y + z 0 First, we rv (x; y; z) = hyz; xz; rg (x; y; z) = hy + z; x + z; x + yi rh (x; y; = h1; 1; 1i Note that rh (x; y; z) 6= 0 If rg (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does not satisfy the constraint g = 0 Assume that rf = rg + rh With the constraints, we have yz = (y + z) + >< xz = (x + z) + xy = (x + y) + xy + yz + xz = 70 >: x + y + z = 0 It becomes >< >: xyz = xy + xz + x xyz = xy + yz + y xyz = xz + yz + z xy + yz + xz = 70 x + y + z = 0 First three equations tell us that < xy + xz + x = xy + yz + y xy + yz + y = xz + yz + z : xy + xz + x = xz + yz + z

15 1 It becomes that < (z + ) (x y) = 0 (x + ) (y z) = 0 : (y + ) (x z) = 0 We notice that if two of the three equations x y = 0, y z = 0 x z = 0 are hold, then we have x = y = z But, by two constraints, it imlies that x = 70 x = 0 There is no such kind of x Thus, we can assume that we can only have at most one of the three equations x y = 0, y z = 0 x z = 0 hold Also, if x, y z are three distinct numbers 6= 0, then we have z = = x = y It cannot haen So, we can assune that x = y, z 6= y, z 6= x Therefore, the constraints becomes x + xz = 70 x + z = 0 By relancing z in the rst equation by z = 0 x, we have x + x (0 x) = 70, or, x 100x + 70 = 0 It tells us that x = = = y the 6 6 results are the same as the rst way 11 #40(b) The lane 4x y + z = intersects the cone z = x + y in an ellise Use Lagrange multiliers to nd the highest lowest oints on the ellise Assume that (x; y; z) is a oint on the ellise To nd the highest lowest oints is equivalent to nd the largest z the smallest z Let f (x; y; z) = z, g (x; y; z) = 4x y + z h (x; y; z) = x + y z First, we calculate rf (x; y; z) ; ; = h0; 0; rg (x; y; z) = h4; ; rh (x; y; z) = hx; y; zi Note that rg (x; y; z) 6= 0 If rh (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does not satisfy the constraint g = 0 Assume that rf = rg + rh With the constraints, we have 0 = 4 + x >< 0 = 4 + y 1 = z 4x y + z = >: x + y z = 0 Note that if = 0, then the rst equations tells us that = 0 But, it will turn the third equation into 1 = 0 which is a contradiction So, we know that 6= 0 Thus, we have < : x = y = z = 1

16 16 This tells us that x = y Thus, the constraints become 7x + z = x z = 0 Here, x z = 0 imlies that z = x If z = x, then the rst equation tells us that x = 7+ In this case, z = x = 7+ If z = x, then the rst equation tells us that x = 7 In this case, z = x = 7 Hence, the highest oint on the ellise is 7+ ; 7+ ; 7+ the lowest oints on the ellise = 0:611 > 1:69 = 7 ; 7 ; 7 because 7+ 7