Multivariable Calculus and Matrix Algebra-Summer 017 Homework 4 Solutions Note that the solutions below are for the latest version of the problems posted. For those of you who worked on an earlier version of the problems I will grade the problems you worked on. First list of problems: 7. We first solve the partial derivatives by standard implicit differentiation. We know Z is a function of x and y. Hence we can write the given equation as x + y + z(x y) = xyz(x y) + x When we differentiate both sides with respect to x we get: x + z(x y) = yz(x y) + yx + (z yx) = yz x + = yz x + z yx When we differentiate both sides with respect to y we get: y + z = xz + xy (z xy) = xz y = xz y z xy We can also solve for the partial derivatives by finding the total differential of the function F (x y z) = x + y + z xyz x (This function can be set to zero to give the original equation). 1
We get the following total differential for F: We set dy=0 and get: df = F F F dx + dy + dz df dx = F + F = 0 We set dx=0 and get: F = F = x + yz + z xy df dy = F + F = 0 F = F = y + xz z xy 15. (a) f x = x y + 4 = 0 which implies that x y = f y = x + y 4 = 0 which implies that x y = This gives us critical points across the line y = x + Now lets check if we have a local maximums local minimums or saddle points. D = f xx f yy f xy = 4 4 = 0 This tells us that we can not tell whether we have local maximums local minimums or saddle points by the second derivative test. (b) f x = e x cos(y) = 0 which implies that y = π + πk1 k1 ɛ Z 6 f y = e x sin(y) = 0 which implies that y = πk k ɛ Z
These equations give us the following. πk = π + πk1 This implies that k-k1=0.5 which is not possible. 6 Hence the function has no critical points and consequently not local maximums local minimums and saddle points. (c) f x = y + 4x = 0 which implies that y = 4x f y = x + 4y = 0 which implies that x 56x 9 = 0 x=0 or x 8 = 1 56 which implies that x = + ( 1 ) and y = + ( 1 ) The critical points are (0 0) ( 1 1 ) and( 1 1 ) D = f xx f yy (f xy ) = 144x y 1 at (0 0)D = 1. Hence we have a saddle point at (00) at ( 1 1) D=8 and f xx( 1 1 ) = > 0. Hence we have a local minimum at ( 1 1 ). at ( 1 1) D=8 and f xx( 1 1 ) = > 0. Hence we have a local minimum at ( 1 1 ). 16. f(x y) = x x + y + y First we will check interior critical points: f x = x = 0 f y = y + = 0 This gives us the coordinates (1.5-1.5). This is outside the prescribed region and hence we conclude that the function has no critical points in the given boundary. Second we check the boundaries of the region:
i)x = 0 : f(0 y) = y + y f(0y) = y + = 0 this gives y=-1.5 and the coordinate (0-1.5) at the corners of this boundary we have: y=0: f(00)=0 y=: f(0)=10 ii)y = : f(x ) = x x + 10 f(0y) = x = 0 this gives x=1.5 and the coordinate (1.5). This is outside the prescribed region and hence we conclude that the function has no critical points in the given boundary. at the corners of this boundary we have: y=: f(1)=8 iii) y = x : f(x x) = 5x + x f(xx) = 10x + = 0 this gives x=-0. and the coordinate (-0.-0.6) which is not the given region. We have already looked all three corner points of the triangular region. We now conclude that the absolute maximum is at (0) where f(0)=10 and the absolute minimum is at (00) where f(00)=0. 17. Let V be the volume and S be the total surface area. We can approach this problem in two ways. Both approaches are consider below: 4
i) Using substitution: From the volume equation we get x = 64 yz S(x y z) = (xy + yz + xz) = ( 64 z S y = (z 64 ) = 0 z = 64 y y S z = (y 64 ) = 0 y = 64 z z 64 + yz + ) = S(y z) y Using the above two equations we get z = 64( 64 z ) which implies that z=0 or z = 64. If z=0 then y is undefined. Therefore the case where z=0 is not a valid critical point of S(yz).When z = 64 z=4. At z=4 we get y=4. Hence (44) is a critical point.as we have seen there are no other critical points for S(yz). Using the volume equation we find that x=4 when z=y=4. Using the second derivative test we find that D = 1 > 0 and S yy = 16 > 0 at (y z) = 4 point (44) is local minimum of S(yz) or in other words (444) is local minimum of S(x(yz)yz). There are no boundaries and only one valid critical point for S(x(yz)yz). Therefore the local min is the absolute min. Hence the surface area is minimal at (xyz)=(444) with S(4 4 4) = 96cm ii) Using Lagrange multipliers: V =< V x V y V z >= λ S = λ < S x S y S z > V (x y z) = xyz = 64 V x = λs X yz = λ(y + z) and rewrite the above equa- V y = λs y xz = λ(x + z) From the equation xyz=64 we get x = 64 yz tion as follows. 64z 18 = λ( + z) yz yz 5
and rewrite the above equa- V z = λs z xy = λ(x + y) From the equation xyz=64 we get x = 64 yz tion as follows. 64y 18 = λ( + y) yz yz Isolating λ we get yz y + z = 64 y(18/yz + z) = 64 z(18/yz + y) Solving for xyand z we get the coordinates (00undefined)and (444). We see that (444) is the only valid critical point and by using the second derivative test or by looking at the fact that the function has not maximal value under the given constraint we conclude that (444) is the point function reaches a minimum. That minimum is given by (4 4 4) = 96cm. Second list of problems: 7. f x = ln(x +y (x + y)(x) )+ x + y f y = ln(x +y (x + y)(y) )+ x + y = (x + y ) ln(x + y ) + (x + y)(x) x + y = 0 = (x + y ) ln(x + y ) + (x + y)(y) x + y = 0 (x + y )ln(x + y ) + x + xy = (x + y )ln(x + y ) + y + xy = 0 (x + y)(x) = (x + y)(y) x = y This implies x = y or x = y 1) x=y: f x (x x) = x (ln(x ))+4x x = 0 This equation leads to x=0 or ln(x ) = which implies x = + 1 e When x=0 y=0 and when x = + 1 e y = + 1 e ) x=-y: f x (x x) = x (ln(x )) x = 0 6
This leads to x=0 or x = + 1 When x=0 y=0 and when x = + 1 y = + 1 Note that at (00) the function is undefined. The critical points are ( 1 1 ) ( 1 1 ) ( 1 e 1 e 1 ) and ( e 1 e ). Now we find D (the Hessian determinant): D = f xx f yy (fxy) = ( x x +y + 6x(x +y ) (x+y)(x) (x +y ) )( x+6y x +y 4y (x+y) (x +y ) ) ( y+x x +y 4xy(x+y) (x +y ) ) at ( 1 ) D = 16 < 0 Hence we get a saddle point at ( 1 1 ) D = 16 < 0 Hence we get a saddle point at ( 1 e 1 e ) D = ( 4 + )( ) < 0 and e e 7 e f xx = ( 4 + ) < 0. hence we get a local maximum at ( 1 e e 7 e 1 e ) at ( 1 e 1 e ) D = ( 4 )( ) < 0 and e e 7 e f xx = ( 4 ) > 0. hence we get a local minimum ( 1 e e 7 e 1 e ) 8. (a) f = λ g where g(x y) = x + y < f x f y >= λ < g x g y > f x = λg x 4 = λ(x) λ = x f y = λg y 6 = λy λ = 1 y The constraint equation gives us: x + y = 14 y = y = + 7
this gives us x = + at f( ) = 14 at f( ) = 14 Therefore we have a max at ( ) and a min at ( ) since we know that these are the points where the function reaches its extreme values under the constraint. (b) f = λ g where g(x y z) = x + y + z < f x f y f z >= λ < g x.g y g z > f x = λg x xy z = λx λ = y z f y = λg y x yz = λy λ = x z f z = λg z x y z = λz λ = x y Combining the three equations of λ and simplifying we get: x = y orz = 0 and x = z or y = 0 The constraint gives us: x + y + z = 1 Using the constraints we get 1 solutions for the conditions set above. This are (0 0 1) (0 1 0)(1 0 0) ( 1 1 1 ) ( 1 1 ) ( 1 1 ) ( 1 ) ( 1 1 1 ) ( 1 1 ) ( 1 ) ( 1 ) The first three coordinates give f=0 which is the minimal value and the rest give f = 1 which is the maximal value. 7 8