Math B6C – Chapter 12 Quiz – SOLUTIONS
* Fall 2001 *
___________________________________________________________________
1. Find
g
 g x for the function g(x, y) = ln(x2 + y2).
x
Then gx(1, 1) = ?
g x  x, y    ln  x 2  y 2 
x
 2
x  y2
 x 2 2  22 x 2
x y
x y
2
Thus, g x 1,1  2 2  1
1 1
___________________________________________________________________


2 g
 g yx . Then gyx(1, 2) = ?
xy
2. For the function g of problem #1, compute
Since this function is infinitely differentiable in the vicinity of (1,2), it follows that its
mixed partial derivatives are equal. Thus gyx = gxy, and so we only need to differentiate
gx (which we found in problem #1) with respect to y:
g xy  x, y   
2x
y x  y 2
 2 x  x2  y2
y
2


1
2
 2 x  x 2  y 2    x 2  y 2 
y
 2 x  x 2  y 2  2 y
g xy  x, y   4 xy 2
2
x
Thus,
2
 y2

g xy 1,2   4 1 22
1
2
 22
8
.

25
___________________________________________________________________
___________________________________________________________________
3. Find the linearization of f (x, y, z) = xz2 – y2cos(z) at (1, 1, ).
L(x,y,z) = ?
  x  1 


L( x, y, z )  f (1,1,0)  f '(1,1,0)   y   1 
  z   0 
    
 x  1
 1  0 2 0  y  1
 z 
 1  2  y  1  1  2 y
___________________________________________________________________
4. For the function h(x, y) = e2x+y, find the best quadratic approximation at
the point (0, 0). Q(x,y) = ?
 x 1
Q ( x, y )  h(0,0)  h '(0,0)     x
 y 2
 x 1
 1   2 1     x
 y 2
 x
y  h ''(0,0)  
 y
 4 2  x 
 
2 1  y 
y  
4 x  2 y 
1
 1  2 x  y   x y  

2
 2x  y 
1
 1  2 x  y   x(4 x  2 y )  y (2 x  y ) 
2
1
Q ( x, y )  1  2 x  y  2 x 2  2 xy  y 2
2
______________________________________________________________________
5. f (x, y, z) = x y 2 – y z + x 2 z . Find the gradient of f at (1,1,1).
The y-component of this gradient is _____ .
f   f x , f y , f z  , and so
f ( x, y, z)   y 2  2 xz, 2 xy  z,  y  x 2 
f (1,1,1)  3,1, 0
______________________________________________________________________
______________________________________________________________________
6. Find the directional derivative of the f in problem #5, in the direction of
at the point (1,1,1).
D1
1,0,1
2
D1 

1,0,1

2
f (1,1,1)  ?
f (1,1,1)  f (1,1,1)
1
1,0,1
2
1
1,0,1
2
1
f (1,1,1) 1,0,1
2
 1 (3,1,0) 1,0,1
2

 3
2
_____________________________________________________________________
7. Find the equation of the tangent plane to x2 – y2 + z2 = 4 at the point ( 1, 1, 2 ).
Let F ( x, y, z)  x2  y 2  z 2 . We seek the tangent plane to the level surface defined by
F ( x, y, z )  4 . One way to do this is to find the linearization of F at the point (1, 1, 2),
which as usual we’ll call L. The tangent plane is then just L(x,y,z) = 4.
  x  1 


L( x, y, z )  F (1,1,2)  F '(1,1,2)   y   1  
  z   2 
   
 x  1
 4  2 2 4   y  1
 z  2
 4  2x  2  2 y  2  4z  8
L( x , y , z )  2 x  2 y  4 z  4
Thus our tangent plane is given by 2 x  2 y  4 z  4  4 , or
x  y  2z  4
_____________________________________________________________________
_____________________________________________________________________
8. Find the linearization of f (x, y, z) = x y z + 1 at (1, 2, 3).
L(x, y, z) =
  x  1 


L( x, y, z )  f (1,2,3)  f '(1,2,3)   y   2  
  z   3 
    
 x 1
 7  6 3 2   y  2 
 z  3 
 7  6  x  1  3  y  2   2  z  3
L( x, y, z )  6 x  3 y  2 z  11
_____________________________________________________________________
9. Find the maximum value of
f (x, y, z) = x + y z
on the sphere x 2 + y 2 + z 2 = 3.
We use the method of Lagrange Multipliers for this problem. Let G( x, y, z)  x 2  y 2  z 2 .
Then extrema of f are to be found at points where, for some , f  G . Then
f  x, y, z   1, z, y  and G  x, y, z    2 x,2 y,2 z  ,
yielding the following equations:
1  2 x
z  2 y
y  2 z
1 . Substitution the second equation into the
2
2
third yields y  2 2 y  4 y , from which we deduce that 4 2  1 , and so the only
1
two possible solutions for  are    , and hence x  1 . So now we know that:
The first of these equations gives us x 
2
x  1
z  y
Now we substitute what we know into the equation G( x, y, z)  3 , since any solution must
be on that sphere:
x2  y 2  z2  3
2
2
 1   y     y   3
2
1  2 y2  3
y2  1
y  1
Now we know that:
x  1
z   y  1
And we must check to see which of these 8 points (since these “plus or minuses” are all
independent) is the maximum:
f ( x, y, z )  x  yz
f (1,1,1)  2
f (1,1, 1)  0
f (1, 1,1)  0
f (1, 1, 1)  2
f (1,1,1)  0
f ( 1,1, 1)  2
f (1, 1,1)  2
f (1, 1, 1)  0
So the maximum value of f on the given sphere is 2 (and the minimum value is –2).
_____________________________________________________________________
10. If a triolic field intensity is described by T(x, y, z) = sin(xyz), then the triolic field is
increasing at (1, 2, /2) most rapidly in the direction of . . .
A scalar field like T increases most rapidly in the direction of its gradient:
T ( x, y, z)   yz cos( xyz ), xz cos( xyz ), xy cos( xyz )
T 1,2, 2    ,  2 , 2 


_____________________________________________________________________
11. How fast is the triolic field increasing in the direction you found in #10?
(approximately)
The length of the gradient vector we just found in #10 is precisely what we seek:
  
2


2
2
2
2 



2   2     4  4
which approximately equals 4.0419061717662000019193482790652316224461245051084 .
_____________________________________________________________________
_____________________________________________________________________
12. Find the tangent plane to the surface defined by z  G( x, y)  x2  e xy at (1,0,0).
The equation is given by z = L(x,y), where L is the linearization of G at (1,0).
 x
1  
 
 
L( x, y )  G (1,0)  G '(1,0)       
y
0
 x  1

 y 
 0  2 1 
 2  x  1  y
L( x , y )  2 x  y  2
_____________________________________________________________________
13. Let T(x, y, z) = x2yz describe the temperature at (x, y, z). How hot is the hottest spot
on the surface implicitly defined by x4  y 4  z 4  1 ?
We use the method of Lagrange Multipliers for this problem. Let
G( x, y, z)  x 4  y 4  z 4 . Extrema of f are to be found at points where, for some ,
T  G . Then
T  x, y, z   2 xyz, x 2 z, x 2 y  and G  x, y, z   4 x3,4 y3,4z3  ,
yield the following equations:
2 xyz  4 x3
x 2 z  4 y 3
x 2 y  4 z3
There are many ways to proceed. One way is to take ratios of the first two equations and
the second two:
2 xyz  4 x3
4
2
3 which yields x   2 y
x z 4 y
x3z  4 y 3 which yields z   y
x 2 y 4 z 3
Now substitute (expressing everything in terms of y) into the constraint equation:
x4  y 4  z4  1
 4 2 y 
4
 y4   y   1
4
2 y4  y4  y4  1
4 y4  1
y 1
2
Then we have y  
1
1
and x   4 . All of these “plusses or minuses” are
2
2
independent, so we must check T out on all these 8 points:
T ( x, y, z )  x 2 yz
T  4 ,
1
 2
1 1 
, 
2 2
1
2 2
T  41 , 1 , 1   1
 2
2
T  41 , 1 ,
 2 2
T  41 , 1 ,
 2 2
2 2 2
1 
1

2 2 2
1 
1

2 2 2
1 1 1 
1
,

 2 2 2 2 2
T  41 , 1 , 1   1
 2 2 2 2 2
T  41 , 1 , 1   1
 2 2 2 2 2
T  41 , 1 , 1   1
 2 2 2 2 2
T  4 ,
Thus the hottest point on the surface has a temperature of
1
2 2
.
_____________________________________________________________________
14. Find the maximum of x + y + z subject to the constraint x2  y 2  xz  1 ?
Suppose that x=0 and y=1. The constraint equation is then satisfied regardless of what z
equals. Thus z can be made as large as we like, and so can the sum x + y + z. This means
there is no maximum of x + y + z subject to the given constraint.
Given any (large) real number R, there are points on the surface x2  y 2  xz  1, the sum
of whose coordinates surpass R.
_____________________________________________________________________
_____________________________________________________________________
15. Find the equation of the tangent plane to the surface x3  2 xy 2  z 2  0 at (1,1,1).
Let F ( x, y, z)  x3  2 xy 2  z 2 . We seek the tangent plane to the level surface defined
by F ( x, y, z )  1 . One way to accomplish this is to find the linearization of F (which as
usual we’ll call L) at (1,1,1). The tangent plane will then be L(x,y,z) = 4.
  x  1 


L( x, y, z )  F (1,1,1)  F '(1,1,1)   y   1 
  z  1 
   
 x  1
 0  1 4 2   y  1
 z  1
 x 1  4 y  4  2z  2
L( x, y, z )  x  4 y  2 z  1
Thus our tangent plane is given by x  4 y  2 z  1  0 , or
x  4 y  2 z  1
_____________________________________________________________________
16. Find any local extrema and saddle points of
f ( x, y)  5x2  4xy  2 y 2  4 x  4 y .
Extrema and saddle points occur at any points where f '( x, y )  0 .
f '( x, y)  10 x  4 y  4 4 x  4 y  4  0 0
This yields the following linear system:
10

 4
4   x   4 
 

4  y   4 
Which has solution x = 0, y = –1. Thus a possible extremum or saddle point is (0, –1).
To see which it is, we invoke the multivariable second derivative test. We need the
second derivative matrix, the Hessian:
 f xx

 f yx
f xy  10 4 


f yy   4 4
The determinant of this matrix is –40 – 16 = –56 < 0, hence our point (0, –1) is a saddle
point.
_____________________________________________________________________
17. Find any local extrema and saddle points of g ( x, y)  x( x  2)sin y over the open
rectangle defined by 0 < x < 3, 0 < y < 3. Give the value of the function at this
point, and classify it as a saddle point, local maximum or local minimum:
Extrema and saddle points occur at any points where g '( x, y )  0 .
g '( x, y)  2( x 1)sin y x( x  2)cos y   0 0
This yields the following nonlinear system:
 2( x  1)sin y  0

 x( x  2)cos y  0
The first equation has solutions x =1 with y=anything, or y=0 with x=anything. The
second equation has solutions x=0 with y=anything, or x=2 with y=anything, or
y   2 with x=anything. The conjunction of the first set of solutions with the second
yields the following possible local extrema or saddle points:
1,  2 ,
0, 0 ,  2, 0
To see whether these points are maxima, minima, or saddle points, we invoke the
multivariable second derivative test. The second derivative matrix, the Hessian, is
 g xx

 g yx
g xy   2sin y
2( x  1)cos y 

.
g yy  2( x  1)cos y  x( x  2)sin y 
The determinant of this matrix is
H ( x, y)  2 x( x  2)sin2 y  4( x 1)2 cos2 y .
Evaluating this Hessian determinant function at each of our candidate points, we get:

H 1,   2,

2
H  0,0   4,
H  2,0   4.
Thus the first point is a possible max or min, while the second two points are saddle
points. To see whether the first point is a max or a min, we evaluate gxx at this point:


g xx 1,   2sin  2  0.

2
2

Since this second derivative is positive, the point 1,  is a relative minimum of g.
2


_____________________________________________________________________
18. Find the absolute maximum of g ( x, y)  x( x  2)sin y over the rectangle defined
by 0  x  3, 0  y  3 . The highest elevation on the graph of z = g(x,y) above
this closed rectangle in the xy-plane is _____ .
In problem #17, we learned that there are no local maximum points within the given
rectangle. Thus we only need to check the edges and corners for any maximum or
minimum points. The four edges can be parametrized as follows:
r1  t   t,0 , 0  t  3
r 2  t    3, t  , 0  t  3
r 3  t    t,3 , 0  t  3
r 4  t    0, t  , 0  t  3
Taking the composition of g over each of these paths, we get:
g  r1  t    g  t,0  0
g  r 2  t    g  3, t   3sin t
g  r 3  t    g t,3  t (t  2)sin3
g  r 4  t    g  0, t   0
Notice that g is identically 0 along the x and y axes.
Taking the derivative (and setting it equal to zero) of the second of these compositions,
we get
d 3sin t  3cos t  0 ,
dt
2n  1  for every integer value of n. The only odd multiple of
2
 that lies within the interval 0 < t < 3 is t   . Thus  3,   could be a maximum


2
2
 2
which has a solution t 
point for g (i.e., the input that yields g’s maximum over the given square in the xy-plane).
Taking the derivative (and setting it equal to zero) of the third of these compositions, we
get
d t(t  2)sin3  sin3 2t  2  0
 

dt
which has solution t = 1. Thus 1, 3 could be a maximum point for g (i.e., the input that
yields g’s maximum over the given square in the xy-plane).
Thus we have the four corners and the above two points upon which to evaluate g to
locate the absolute maximum of g over the given square.
g ( x, y )  x( x  2)sin y
g (0,0)  0
g (3,0)  0
g (0,3)  0
g (3,3)  3sin3  .423360024179601
g  3,    3

2
g (1,3)   sin3  .141120008059867
The maximum of g over the given square is thus 3, the highest elevation of g over the
square.
_____________________________________________________________________
19. Find the absolute minimum of g ( x, y)  x( x  2)sin y over the rectangle defined
by 0  x  3, 0  y  3 . The lowest elevation on the graph of z = g(x,y) above this
closed rectangle in the xy-plane is _____ .
All we need do is to compare the minimum found in problem #17 with the values just
tabulated in problem #18:

g 1,   1
 2
Comparing this with the previous values, we see that the relative minimum of g is also
the absolute minimum over the given square.
_____________________________________________________________________
20. How many local extrema and saddle points has h( x, y)  x3  y3  3x2  3 y 2  8 ?
Such points occur when h '( x, y )  0 , i.e. when
h '( x, y )  3x 2  6 x 3 y 2  6 y   0 0 .
This yields the following nonlinear system:
 3x 2  6 x  0
,
 2
3
y

6
y

0

which simplifies to
 x( x  2)  0
.

 y ( y  2)  0
The first equation has solutions x = 0 with y=anything, or x = –2 with y=anything. The
second equation has solutions y = 0 with x=anything, or y=2 with x=anything. The
conjunction of the first set of solutions with the second yields the following possible local
extrema or saddle points:
0, 0 , 0, 2  ,  2, 0 ,  2, 2  .
To see whether these points are maxima, minima, or saddle points, we invoke the
multivariable second derivative test. The second derivative matrix, the Hessian, is
 hxx

 hyx
hxy  6 x  6
0 

.
6 y  6
hyy   0
The determinant of this matrix is
H ( x, y )   6 x  6 6 y  6  36  x  1 y  1 .
Evaluating this Hessian determinant function at each of our candidate points, we get:
H  0,0  36, H  0,2   36, H  2,0   36, H  2,2   36.
Thus (0, 0) and (–2, 2) are both saddle points (since the determinant is negative for these
points). To see whether the remaining two points are max or min, we evaluate hxx at
these points:
hxx  0, 2   6  0 and hxx  2, 0   6  0
and so (0, 2) yields a relative minimum and (–2, 0) yields a relative maximum of h.
So h has two saddle points, one local minimum and one local maximum point; in total h
has 4 local extrema and saddle points.
_______________________________________________________________________