Math 63, Practice Exam II, Solution. (a) f =< f s, f t >=< s e t, s e t >, an v v = <,>, so D v f(, ) =< ()e, e > <, > =< 4, 4 > <, > = 4. (b) f =< xy 3, 3x y 4y 3 > an v =< cos π, sin π >=<, 4 4 >, so D v f(, ) =< 4, 8 > <, >= 6. (c) f =<, x, x > an v y+z (y+z) (y+z). (a) f =< y x, y x = <,,3> v 4, D v f(4,, ) =<,, 3 >,, > < = 9 4 4. >, then f(, 4) =< 4, 4 >. Therefore the maximum change of rate is f(, 4) = 4 an it occurs in the same irection of f(, 4) =< 4, 4 >. (b) f =< 4x 3 y 3 z, 3x 4 y z, x 4 y 3 z >, then f(,, ) =< 4, 3, >. Therefore the maximum change of rate is f(,, ) = 9 an it occurs in the same irection of f(,, ) =< 4, 3, >. 3. (a) By simplifying f x = xe y x + (x + y )( x)e y x = we have f y = ye y x + (x + y )(y)e y x = e y x x( x y ) = e y x y( + x + y ) = x = or x y = y = So we have the following solutions i. x =, y = ii. Solve x y = y = gives two points: (, ) an (, ). x( x y ) = y( + x + y ) =
Combine all together, there are three critical points (, ), (, ) an (, ). To classify the critical points, we nee to calculate Since It is easy to see that D = f xx f yy f xy f xx = e y x (( x y )( x ) x ), f xy = 4xye y x (x + y ), f yy = e y x (( + x + y )( + y ) + y ). D(, ) = 4 >, f xx (, ) = >, D(, ) = 6e <, D(, ) = 6e <. So f(, ) = is a local minimum an (, ), (, ) are sale points. (b) First, we have f x = 6x + y + x =, f y = xy + y = By solving f y = xy + y = y(x + ) =, we have either y = or x =. i. If y =, substituting it into f x = 6x + y + x = gives x = or x = 5/3. ii. If x =, substituting it into f x = 6x + y + x = gives y = or y =. Combine the above, there are four critical points: (, ), ( 5/3, ), (, ), (, ). Next we nee to calculate D. Since It is easy to see that f xx = x +, f xy = x +, f yy = y. D(, ) = >, f xx (, ) = >, D( 5/3, ) >, f xx ( 5/3, ) <, D(, ) <, D(, ) < Hence f(, ) = is a local minimum, f( 5/3, ) = 5/7 is a local maximum, an (, ), (, ) are sale points.
4. Step : Fin the local minimums an maximums. f x = 4 x = f y = 6 y = x = y = 3 Then f(, 3) = 3 is a caniate for local minimums an maximums. Step : fin minimums an maximums on the bounary eges. On the bottom bounary y =, we have f(x, ) = 4x x an it has minimum at f(, ) = f(4, ) = an maximum at f(, ) = 4. On the left bounary x =, we have f(, y) = 6y y an it has minimum at f(, ) = an maximum at f(, 3) = 9. On the right bounary x = 4, we have f(4, y) = 6y y an it has minimum at f(4, ) = an maximum at f(4, 3) = 9. On the top bounary y = 5, we have f(x, 5) = 4x x +5 an it has minimum at f(, 5) = f(4, 5) = 5 an maximum at f(, 5) = 9. Compare step an, we have the absolute minimum at f(, ) = f(4, ) = an the absolute maximum at f(, 3) = 3. 5. The istance is given by f(x, y, z) = (x 8) + (y ) + (z 8) an the constraint is given by g(x, y, z) = 8x y + 4z = 6. To avoi long notations, we enote = (x 8) + (y ) + (z 8), then f =< x 8, y, z 8 g =< 8,, 4 > >, Using the Lagrange multiplier metho, we have x 8 = 8λ y = λ Noticing that z 8 = 4λ 8x y + 4z = 6 ( x 8 ) + ( y ) + ( z 8 ) =, Therefore (8λ) + ( λ) + (4λ) =, which implies λ = ±/ 8. Now, we have x = 8λ + 8, y = λ +, z = 4λ + 8 substitute them into the constraint g(x, y, z) = 8x y + 4z = 6, we have 64λ + 64 + λ + 6λ + 3 = 6 an hence 8λ =. Since we know that λ = ±/ 8, it is easy to see that = ±/ 8. Finally, recall that we efine = (x 8) + (y ) + (z 8), which is exactly the istance. So at this point we alreay have the answer that the shortest 3
istance is = / 8. Of course one can solve for (x, y, z) by plug in values of λ an back into x = 8λ + 8, y = λ +, z = 4λ + 8 6. Let x, y, z be the length of three sies. The volume is given by f(x, y, z) = xyz an the constraint is surface area g(x, y, z) = xy + yz + xz = 5. Using the Lagrange multiplier we have yz = λ(y + z) xz = λ(x + z) xy = λ(x + y) xy + yz + xz = 5 Multiplying the first equation by x, the secon equation by y an the thir equation by z, we have xyz = λ(xy + xz) xyz = λ(xy + yz) xyz = λ(xz + yz) λ(xy + xz) = λ(xy + yz) = λ(xz + yz) λ = or (xy + xz) = (xy + yz) = (xz + yz) It is not har to see that λ can not be since otherwise one of x, y, z has to be zero. Now solving (xy + xz) = (xy + yz) = (xz + yz) gives x = y = z. Substitute it into the constraint xy + yz + xz = 5 gives x = y = z = 5. 7. The approximation is 8. (a) i= 6 f(x i, y j ) A = 8f(4, ) + 8f(8, ) + 8f(4, 4) + 8f(8, 4) = 86 j= 3 (x+y)3/ x= x= y = 6 [ 3 (y+)3/ 3 y3/ ] y = 4 5 65/ 4 5 65/ 4 5 5/. (b) 9 x 3y 9 y x = x + 3y / x x + y= y= x = 9 = 3 4 ln(x + ) 9 = 3 ln 8 4 3x/ x + x 4
(c) first change the orer of the integral: x x + y xy = = Using integration by parts, we have ln(y + 4)y = y ln(y + 4) Similarly, one can calculate that ln(x + y ) x= x= y [ ln(y + 4) ln(y + )]y y y + 4 y ( 4 y + 4 )y y y y + 4 y [y tan y ] [ tan ] + tan ln(y + )y = ln + tan = ln + π 4 Combine the above, we have the final result x ln 5 x + yy x = + tan ln π 4. () We woul like to first shange the orer of the integral. Since so we have y x3 + xy = D = (x, y) y, y x } = (x, y) x, y x } = x x3 + y x = ( x 3 + )y y=x y= x x x 3 + x = 9 (x3 + ) 3/ = 9 ( 8 ). 5