Math 6
Spring 2008
February Test
Directions Solve the following problems showing all work clearly and in a wellorganized fashion. Use Lagrange multipliers to find the constrained extrema.
1. Find the extreme values of f ( x, y)  x 2  2 y 2 on the disk x 2  y 2  1 .
Check the interior: f x  2 x and f y  4 y , so the only critical point is the origin,
and f(0,0) =0.
Check the boundary:
f ( x, y )  x 2  2 y 2
g ( x, y )  x 2  y 2  1
Hence,
f  2 x, 4 y 
g  2 x, 2 y 
So the three equations are
2 x  2 x
4 y  2 y
x2  y 2  1
From the first of these, either x  0 or   1 . If x = 0 then the third equation gives
y  1 . If   1 then the second equation requires that y = 0, in which case the third
equation requires that x  1 .
Hence, the possible extrema are  0, 1 and  1,0  . Evaluating the function at all
four candidate points, we find
f (0,1)  2
f (0, 1)  2
f (1,0)  1
f ( 1,0)  1
And hence the max value is 2 and the min value is 1.
Combined Since the value of f at the only critical point is 0, which is less than the
constrained minima, the minimum of the function is 0 and the maximum is 2.
2. Find the maximum value of the function f ( x, y, z )  x  2 y  3z on the curve of
intersection of the plane x – y + z = 1 and the cylinder x 2  y 2  1
g ( x, y , z )  x  y  z  1
Let
h ( x, y , z )  x 2  y 2  1
Then
f  1, 2,3 
g   ,  ,  
h  2 x, 2  y,0 
And so, for f  g  h ,
1    2 x
2    2  y
3
Combining the last and first equations, 2  x  2 , so x  
and second equations gives y 
1

. Combining the third
5
.
2
29
29
25
and   
.
 1 , so  2 
2
4
2
 4
2
x
29
5
Thus, y  
29
7
z  1 x  y  1
29
2
5  
7 

 2 
So, f 
  3 1 
  3  29
29
29  
29 

Hence,
1
2

And so the max is 3  29
3.
In finding the largest value of a four-variable function subject to three
constraints, how many equations in how many unknowns would the Lagrange
method require?
7
Where specifically do the various needed equations come from?
From the four components of the gradient of the four-variable function and the three
constraint conditions.
Bonus: What’s the equation of the curve of intersection of the two surfaces in the second
problem?
 cos t,sin t,1  cos t  sin t  .
If you use Mathematica’s ParametricPlot3D with the
previous argument and let t run from 0  2 , you get an ellipse that looks like the
appropriate curve of intersection between the unit cylinder and the given plane:
Y
1.0
0.5
1.00.50.0
0
1
2
1.0
0.5
0.0
X
0.5
1.0
Z