Math 51 TA note class 18
May 26, 2010
chap 13 Extrema of multivariable functions
Example 1. Find the absolute extrema of f (x, y) = sin x cos y on the rectangle
{(x, y) | 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2π}
Answer.
Proposition 1 (Lagrange Multipliers). Constraints: g1 (x) = c1 , g2 (x) = c2 , . . . , gk (x) =
ck , f : Dn → R is real-valued function. Let
S = {x ∈ X | g1 (x) = c1 , g2 (x) = c2 , . . . , gk (x) = ck }.
If f on S has an extremum at at a and
{∇g1 (a), ∇g2 (a), . . . , ∇gk (a)}
are linear independent, then there exist scalar λ1 , λ2 , . . . , λk such that
∇f (a) = λ1 ∇g1 (a) + λ2 ∇g2 (a) + . . . + λk ∇gk (a) .
Remark. If {∇g1 (a), ∇g2 (a), . . . , ∇gk (a)} are linear dependent at some point a , then
a can be another critical point but we can’t use Lagrange multipliers at a . We need to
find other method, for example, compute directly.
Example 2. f (x, y) = 2x2 + y 2 − y + 3 on a unit disk x2 + y 2 ≤ 1. Find the global
extrema of f.
∂f
∂x
∂f
∂y
= 2y − 1. Set
4 0
∂f
∂f
∂x = ∂y = 0 we find the only critical point is (0, 1/2). Hf (0, 1/2) = 0 2 . So
(0, 1/2) is a local min. Furthermore, this point is indeed in the unit disk.
(2) Second, we find the max and min on the boundary. The boundary is the unit
circle and it satisfies the equation x2 + y 2 = 1. Therefore x2 = 1 − y 2 . Plug this
back we get
Answer.
(1) First, we find the critical points of f.
= 4x,
f (x, y) = 2(1 − y 2 ) + y 2 − y + 3 = 5 − y 2 − y .
0=
∂(5−y 2 −y)
∂y
= −2y − 1. Therefore y = −1/2, and use x2 + y 2 = 1 we conclude
two possible extremum are (±
√
3
1
2 ), − 2 ).
√
And f (±
3
2 )
= 5 41 .
By (1) and (2) we conclude (0, 1/2) is the global min and (±
global max.
√
3
1
2 , −2)
are the
Example 3. Find critical points of f (x, y, z) = xyz on the surface 2x + 3y + 6z = 6.
Answer. Let g(x, y, z) = 2x + 3y + 6z. ∇f = (yz, xz, xy), ∇g = (2, 3, 6) By Lagrange
multiplier, we have
yz = 2λ, xz = 3λ, xy = 6λ .
Solve x, y, z in terms of λ and plug it back in 2x + 3y + 6z = 6, we get (x, y, z) =
(1, 2/3, 1/3).
1
2
Example 4. Use Lagrange multipliers to prove that the distance from a point (x0 , y0 , z0 )
Ax0 + By0 + Cz0 + D
to the plane Ax + By + Cz + D = 0 is
A2 + B 2 + C 2
Example 5. Find the point closest to the point (2, 5, −1) and on the line of intersection
of the planes x − 2y + 3z = 8 and 2z − y = 3.
Answer. There are two ways to do this problem:
(1) First way. The intersection of the planes is a line. If we solve 
thesystems
 of
1
2
linear equations x − 2y + 3z = 8 and 2z − y = 3, the solution is {t 2 + −3},
1
0
   
1
2
for any real number t. Therefore the distance of {t 2 + −3} and (2, 5, −1)
1
0
p
2 . To find the point closest to the point (2, 5, −1)
is t2 + (2t − 8)2 + (t + 1)p
is equivalent to minimize t2 + (2t − 8)2 + (t + 1)2 , and then is equivalent to
minimize t2 + (2t − 8)2 + (t + 1)2 .
d 2
[t + (2t − 8)2 + (t + 1)2 ] = 2t + 4(2t − 8) + 2(t + 1) = 12t − 30
dt
. Therefore the critical point is t = 52 . And the point is ( 29 , 2, 52 ).
(2) Second way. Let f (x, y, z) = (x − 2)2 + (y − 5)2 + (z + 1)2 . f is minimized if
the distance from (x, y, z) to (2, 5, −1) is minimized. And two constraints are
g1 = x − 2y + 3z = 8, g2 = 2z − y = 3. ∇f = λ1 ∇g1 + λ2 g2 implies


 
 
2(x − 2)
1
0
2(y − 5) = λ1 −2 + λ2 −1 .
2(z + 1)
3
2
Solve x, y, z in terms of λ1 , λ2 and plug in the two constraints, we can get
(x, y, z) = ( 92 , 2, 25 )
Example 6. Heron’s formula for the area of a triangle whose sides have length x, y, z is
Area=[s(s − x)(s − y)(s − z)]1/2 where s = (x + y + z)/2. Use Herons formula to show
that for a fixed perimeter 2s, the triangle with the largest area is equilateral.
Answer. Let f (x, y, z) = (Area)2 = s(s − x)(s − y)(s − z). f has max at a point if and
only if the area has max at the same point. The constraint is x + y + z = 2s. Using
Lagrange multiplier, we get
−s(s − y)(s − z) = λ, −s(s − x)(s − z) = λ, −s(s − x)(s − y) = λ
Therefore −s(s − y)(s − z) = −s(s − x)(s − z). By the triangle inequality, s − x =
(y + z − x)/2 > 0. Similarly, s − y > 0, s − z > 0. So s − y = s − x, x = y.
Similarly y = z. We conclude the triangle is equilateral. It is a max because if we take
other (x, y, z) satisfies x + y + z = 2s, to make it easy, pick (s, s, 0) then the area is 0,
definitely strictly smaller than area of a equilateral triangle with perimeter 2s.
Example 7. A cuboid box is to be made to contain 1 L of oil. Find the dimensions that
will minimize the cost of the metal to manufacture the box.