( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) Math 3435 Homework Set 4 Solutions 10 Points. 1. (2 pts) First we need the gradient

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1 Math Homework Set Solutions 10 Points 1. ( pts) First we need the gradient 6 F ( x,, z) = 6 + z + x x z 6 F = xz,, x + z + z F, 1, = 16,1, r t =, 1, + t 16,1, = 16 t, 1 + t, t 16 x z = 0 16x + z =. ( pts) First get all the derivatives and D. f = 6x + x + x 7 f = x 8 Find the critical points.. 6x + x + x 7= 0 x f = x + + f = x f = x xx 1 x 8 Dx, = 1x+ + x 8 16x x 8 = x = 0 = 0, x =± ± = 0 : 6x + x 7 = 0 x= = 1.67, x= : + = 0 =± x= : + 9 = 0 =± Finall, classif them. D( 1.67,0) = 0.9 f xx ( 0, 1.096) = 1.6 relative maximum ( ) ( ) = D ,0 = 91.8 saddle point D = 1 saddle point D, 1 saddle point. ( pts) First we need to find all the critical points. h = 10 + x h = 10x+ x x ( 10 + x)( ) = 0 10x+ x = 0 x=, x= 0, = x= : 10 = 0 = 0, x= 0 : 10 = 0 = 0, = : 10 + x = 0 x= We have a large number of critical points :

2 Math Homework Set Solutions 10 Points (,0, ) (,,0,0,0,, ) (,) From a quick sketch of D we can see that the three sides are given b, Bottom : = 0, x 0 Left : x=,0 8 Hp. : = x, x 0 From this we can see that onl (, ) and (0,0) lie in the region or on the boundar as (0,0) does. Now let s work through each side. Bottom : f ( x) h( x ) f ( ) =,0 = 0 = 0 No critial points So, from the bottom we get the end points (0,0) and (-,0). Note that the function will in fact be zero along the bottom of the region at ever point! Left : f = h, = 8 f = 8 = This point in is the range of for the left and so we get the critical point (-,) and the endpoints (-,0) and (-,8). Hpotenuse : f x = h x, x = 10x+ x x + 8x f x = 8 x + 1x + 0 x x=,, 0 We get three critical points on the hpotenuse (0,0), (-, 8) and (, ) along with the two points (0,0) and (-,8) from the hpotenuse (both of which are also in the critical point list ). Finall, all we need to do is plug in the various points and identif the absolute extrema. h, = 0 h 0,0 = 0 h,0 = 0 ( ) h, = h,8 = 6 h, = = So, it looks like the absolute maximum is 0 which occurs at minimum is -6 which occurs (-,8)., and the absolute 7. ( pts) In this case all three terms of the constraint are guaranteed to be positive (because each is squared) and so x, and z must have limits on them and so absolute extrema must exist for this problem. 16x= 8λx x= 0 or λ = + + = x z 1 1 λ ( λ ) 0= λ = 0 0 = λz z =

3 Math Homework Set Solutions 10 Points Note that from the third equation we can see that both z and (more importantl) λ cannot be zero. This also means that, from the second equation, our onl option for is to have = 0. The first equation has two possibilities so we just need to go through each. x = 0: So, we know the values of both x and so use the third equation to find λ and hence the value of z ( 0) + = = 1 λ λ λ =± 1 z =± 1 0,0, 1. This ields the two points ( 0,0,1 ) and λ = : In this case we know the value of and z (from the third equation) so plug these into the constraint to find the value(s) of x. 1 1 x = x + = 1 x=± =± We then have the two points ( 0.01, 0, 0.) and ( 0.01, 0, 0.). Finall, we just need to plug these values into the function to identif the absolute extrema. f 0, 0,1 = f 0, 0, 1 = f ± 0.01, 0, 0. =. The maximum is then which occurs at( 0,0,1 ) and the minimum is -. which occurs at the two points ( ± 0.01, 0, 0.). Not Graded. We need to find points on the surface for which the normal vector will be parallel to the normal vector from the plane 1, 8,. We know from Calculus II that two vectors will be parallel if one is a scalar multiple of the other. So we re looking for points where the gradient vector for the surface will be parallel to the normal vector from the plane. Or, x,, 1z = c 1,8, = c,8 c, c x= c, = cz, = c 6 Now, plug these into the equation of the surface to determine values of c. 18 ( 1 ) ( 1 ) 6 c + c 6 c = 1 c = c=± 18 =± 0.60 So, it looks like we have two points (one for each c) that will work, ,1.08, , 1.08, 0.101

4 Math Homework Set Solutions 10 Points. Note that I had a tpo problem on this and should have added the condition that > 0 to avoid the obvious issues at = 0 in g and of course in D. Sorr about that! First get all the derivatives and D. x g = + x 0x+ g = 1 x 6 1 x xx = 6 0 x = = (, ) = x x g x g g Dx Find the critical points.. x 1 = 0 = 6x 6 ± 6x + x 0x+ = x 8x+ = 0 x = = 0.1,.1 7 x= 0.1: =.7090 = 7.87 x=.1: = 1.9 = Finall, classif them. D( 0.1, 7.87) = f xx ( 0.1, 7.87) = 17.9 D(.1, ) = ,7.87 is a saddle point. So, ( 0.1,7.87 ) is relative maximum and 6. First notice that because both the x and are raised to the th power both terms in the left side of the constraint are positive and so neither x or can be too large (i.e. the must have limits on them namel : x, ) and so absolute extrema must exist for this problem. 16x= λx = λ x + = Note that if λ = 0 then both x and are also zero and this violates the constraint. So λ 0. From the first and second equations we get, λx 16x= x( λx ) = 0 x= 0, x = λ 1 λ = ( λ 1) = 0 = 0, = λ If x=0 we get that =± =± from the constraint so the first two points we need are ( 0, ) and ( 0, ) points are (,0 ) and (,0). Likewise if =0 we get x =± =± so the next two.

5 Math Homework Set Solutions 10 Points Since we have now dealt with the cases of x=0 and =0 we can now assume that x 0 and 0. In this case we have the remaining case from each equation. So, plug the final two conditions into the constraint to get, + = = λ = λ =± λ λ λ Note that if we use the negative λ we will get complex x and so we can ignore that one. Use the positive values gives, x x = 1.98 =± = 0.81 =± Because these all came from the same value of λ we can mix/match these to get the four points , 0.981, 1.087, 0.981, 1.087, 0.981, 1.087, Therefore, all we need to do is plug in the eight points from above into the function and note that signs on the x and won t change the answer both x and are squared. f ( 0, ± ) = f ( ±, 0) = 16 f ( ± 1.087, ± 0.981) = So, the minimum of occurs at the two points ( 0, ± ) while the maximum of occurs at the four points ( ± 1.087, ± 0.981). 8. In this case all three terms of the constraint are guaranteed to be positive (because each is squared) and so x, and z must have limits on them and so absolute extrema must exist for this problem. z = λx xz = 8λ x = 6λz x + + z = 6 Multipling the first equation b x, the second b and the third b z gives, xz = λx xz = 8λ xz = 6λz Setting the first and second equal as well as the first third equal gives that, λ x = 0 λ = 0 or x = λ x z = 0 λ = 0 or x = z Let s start off b assuming that λ = 0. In this case the three original equations become,

6 Math Homework Set Solutions 10 Points z = 0 = 0 or z = 0 xz = 0 x = 0 or z = 0 x = 0 x = 0 or = 0 Now, we can t have all three zero since that won t satisf the constraint. However, notice that if =0 then we must have either x or z be zero in order to satisf the second equation. Likewise if x=0 then either or z must be zero to satisf the first equation. Finall, if z=0 then either x or must be zero in order to satisf the third equation. So, we can t have all three be zero and we can t have onl one be zero. However, in this case, we can have two of them be zero. So, if we assume that two are zero and plug these into the constraint to solve for the third we get the following six points, ( 0,0, ± 1 ) ( 0, ±,0) ( ± 6,0,0) 1 1 Now, let s assume that λ 0. This forces = x and z = x. Plugging these into the constraint gives, x + x + x = x = 6 x=± Plugging this into = x and z = x we can see that we also have = ± and z = ± and that we can have an combination of these in the coordinates and so we will get the following eight points, ( 1,, ),( 1,, ),( 1,, ),( 1,,) ( 1,, ),( 1,, ),( 1,, ),( 1,, ) Now plugging into the function isn t as bad as it looks. Let s start with these eight points. Notice that we ll get ± 1 for ever point and the sign will depend upon the number of minus signs in each term. Likewise, the first six points will all give 0 in the function. So, the absolute maximum is 1 which occurs at, ( 1,, ),( 1,, ),( 1,, ),( 1,, ) and the absolute minimum is 1 which occurs at, ( 1,, ),( 1,, ),( 1,, ),( 1,, )