Thursday, October 4 Solutions Constrained Min/max Via Lagrange Multipliers
Thursday, October 4 Solutions Constrained Min/max Via Lagrange Multipliers
Thursday, October 4 Solutions Constrained Min/max Via Lagrange Multipliers
(b) Is C bounded?
SOLUTION:
No. Given arbitrarily large y values we can find an x value which satisfies the equation. To
p
3
see this notice that y = 16 x 3 , so we can input arbitrarily large (or small) x values and
get a y value for that input.
(c) Is C closed?
SOLUTION:
Yes, C is closed in R2 .
2. Consider the function f (x, y) = e x y on C .
(a) Is f continuous? What does the Extreme Value Theorem tell you about the existance of
global min and max of f on C ?
SOLUTION:
Yes, f is continuous. Since C is not bounded, the Extreme Value Theorem does not tell you
anything about the existence of a global min and max of f on C .
(b) Use Lagrange multipliers to determine both the min and max values of f on C .
SOLUTION:
Let g (x, y) = x 3 + y 3 . Our constraint is g (x, y) = 16. f = (ye x y , xe x y ) and g = (3x 2 , 3y 2 ),
so using the method of Lagrange multipliers we need to find simultaneous solutions in x
and y of the following three equations:
x3 + y 3 =
xy
16
= 3x
(1)
2
(2)
xe x y = 3y 2
(3)
ye
(b) Use Lagrange multipliers to find the points on S that are closest to (4, 2, 0).
SOLUTION:
Minimize the square of the distance function D = (x 4)2 + (y 2)2 + z 2 from the point
and g = 2x, 2y, 2z . From the picture it is clear that D attains a global minimum value
on S (i.e. there are points which are closest to (4, 2, 0)). So one of the critical points we find
using Lagrange multipliers will correspond to this minimum value and we simply need
to evaluate D at each of the critical points and take the smallest to find the minimum
distance. Using the method of Lagrange multipliers we get the system (divide out by 2
first):
(x 4) =x
(y 2) =y
z = z
If 6= 1 then z = 0 from the last equation so the constraining equation z 2 = x 2 +y 2 implies
that x = y = 0. If = 1 then the top two equations give x = 2 and y = 1. So the three
p
p
possible points of minimum distance from (4, 2, 0) are (0, 0, 0), (2, 1, 5), and (2, 1, 5). By
calculation we see that the squares of the distances of each of these from (4, 2, 0) are 20, 10,
p
p
and 10 respectively. So the two points (2, 1, 5) and (2, 1, 5) on the cone z 2 = x 2 + y 2 are
of minimum distance from the point (4, 2, 0).
4. For the function shown on the back of the sheet, use the level curves to find the locations and
types (min/max/saddle) for all the critical points of the function:
f (x, y) = 3x x 3 2y 2 + y 4
Use the formula for f and the 2nd -derivative test to check your answer.
SOLUTION:
Mins and maxes occur where the level curves shrink toward a point and saddle points occur
where the level curve intersects itself. From looking at the set of level curves it appears that
f (x, y) has minimums at (1, 1) and (1, 1), a maximum at (1, 0), and saddle points at (1, 0),
(1, 1), and (1, 1).
Now lets find the critical points precisely. f x = 3(1 x 2 ) and f y = 4y(y 2 1). So f has critical
points at (1, 0), (1, 1), (1, 1), (1, 0), (1, 1), and (1, 1). f xx = 6x, f y y = 12y 2 4, and f x y = 0,
so the Hessian is D(x, y) = f xx f y y ( f x y )2 = 6x(12y 2 4). D(1, 0), D(1, 1), and D(1, 1) are all
negative, so these are saddle points. D(1, 0), D(1, 1), and D(1, 1) are all positive so these are
maxes and mins. f xx (1, 0) < 0 so (1, 0) is a local max. f xx (1, 1) and f xx (1, 1) are both positive
so these are local mins. This analysis agrees with our guesses.
5. If the length of the diagonal of a rectangular box must be L, what is the largest possible volume?
SOLUTION:
Set x =length of the box, y = width of the box,z =height of the box. This simply supposes that
the box is sitting in the octant x 0, y 0, and z 0 with its edges along each axis. The volume
function is then V = x y z and the constraint is that L 2 = x 2 + y 2 + z 2 . Using the method of
Lagrange multipliers we get the system of equations:
y z =2x
xz =2y
x y =2z
x 2 + y 2 + z 2 =L 2
Since we want to maximize volume we can assume that x > 0, y > 0, and z > 0. This rules out
the possibility = 0 (since = 0 implies at least two of the variables x, y, and z are 0). Also this
means we can multiply the first equation by x, the second by y, and the third by z to get a new
system:
x y z =2x 2
x y z =2y 2
x y z =2z 2
This implies that x 2 = y 2 = z 2 . Coupling this with the constraints x > 0, y > 0, z > 0 we see that
this means x = y = z. Plugging this into the constraining equation L 2 = x 2 + y 2 + z 2 we get that
p
p
p
L 2 = 3x 2 or x = L/ 3. So V = (L/ 3)3 = L 3 /(3 3) is the biggest possible volume for the box.