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4.7 Maximum and Minimum Values (11.7)
Def : A function of two variables z=f(x,y) has a local maximum at (x,y)=(a,b) if f ( x , y ) ≤ f ( a , b )
when (x,y) is near (a,b). In other words there exists a small disc D centered at (a,b) such that for
all ( x , y ) ∈ D the inequality f ( x , y ) ≤ f ( a , b ) holds.The value f(a,b) is called local maximum value.
Similarly, f(a,b) is a local minimum if there is a disc centered at (a,b) such that f ( x , y ) ≥ f ( a , b )
for all ( x , y ) ∈ D.
Thm (Fermat’s): If z=f(x,y) has a local maximum or minimum at (a,b) and the first order partial
derivatives of f exists, then fx ( a , b ) = 0 and fy ( a , b ) = 0.
Note : This also hints that max/min could be at points where at least one of partial derivatives
doesn’t exists.
Second Derivative Test : Suppose that z=f(x,y) has continuous partial derivatives on a disc
centered at (a,b) and suppose (a,b) is a critical point, that is fx ( a , b ) = 0 and fy ( a , b ) = 0. Define
D = D (^) ( a , b ) = fxx fyy − (^) ( fxy ( a , b ))
2
If D > 0 and fxx ( a , b ) > 0 then f ( a , b ) is a local minimum.
If D > 0 and fxx ( a , b ) < 0 then f ( a , b ) is a local maximum.
If D < 0 , then f ( a , b ) is a saddle point.
If D^ =^0 -‐ the test is indecisive use another method.
Note : It may be easy to remember formula of D it with use of Hessian Matrix
H 2 × 2 =
fxx fxy
fyx fyy
fxx fxy
fxy fyy
, since D = det H.
Def : An analogous term to a closed [ a , b ] /open ( a , b ) /clopen ( [ a , b ) or ( a , b ] ) interval in is a
closed/open/clopen set in 2 (also in n ( ) ).^ A set is^ closed if it contain all it’s boundary
points.
Note : A domain is considered an open set.
Def : A set S is a bounded set if there is a disc D such that S ⊂ D.
Thm : Extreme Value Theorem for function of two variables z=f(x,y). If f is continuous on a
closed bounded set D ⊂ 2 then f attains its absolute maximum and minimum values.
Alg : To find the abslolute minimum and maximum values of a continuous function f on a closed
bounded set D :
1) Find the values of f at the critical points of f in D.
2) Find the extreme values of f on the boundary of D.
3) The largest/smallest value from 1,2 is the absolute max/min.
Ex 8. Find and classify all critical points of f ( x , y ) = ( x − y ) e
xy
in
2
The function is differential everywhere (as a product of differential functions), so the critical points are only at zeros of partial derivatives.
thus x = − y = ± 1 / 2 and a critical points are ( 1 / 2 ,− 1 / 2 ), ( − 1 / 2 , 1 / 2 ).
Calculate D = fxx fyy − fxy 2 :
fxx = ye
xy ( 2 + xy − y
2 ) fyy = xe
xy (− 2 + x
2 − xy ) fxy = e
xy ( 2 x + x
2 y − xy
2 − 2 y )
D = yxe
2 xy ( 2 + xy − y
2 )(− 2 + x
2 − xy ) − e
2 xy 2 x + x
2 y − xy
2
( −^2 y )
2
D ( x ,− x ) = − x
2 e
− 2 x^2 ( 2 − x
2 − x
2 )(− 2 + x
2
2 ) − e
− 2 x^2 2 x − x
3 − x
3
( +^2 x )
2
= 4 x 2 e − 2 x^2 1 − x 2
2 − 2 − x 2
2
( ) =^4 x
2 e − 2 x^2 2 x 2
( −^3 )
x = 1 / 2
= 2 e − 1
(^1 −^3 ) =^ −^4 e
− 1 < 0
Thus we got a saddle point.
Ex 9. Find absolute min/max of f ( x , y ) = x
2 − xy + y
2
in rectangular − 1 ≤ x , y ≤ 1
The critical points only at zeros of partial derivatives (why?):
f ' x = 2 x − y = 0
f ' y = − x + 2 y = 0
⇒ 2 y = x = y / 2 ⇒ ( x , y ) = ( 0 , 0 )
D = fxx fyy − fxy
2
= 2 ⋅ 2 − (− 1 )(− 1 ) = 3 > 0 and also f '' xx = 2 > 0 therefore we got a local minimum at
Next we look at boundaries we have
f (± 1 , y ) = ( ± 1 )
2 ± y + y 2
f ( x ,± 1 ) = x
2
+ x + ( ± 1 )
2
which are smiling parabolas (second derivative =2>0) centered (with min) at ± 12 respectively
and a minimum value is 34. This value is a local minimum on the boundary, but is above the
minimum inside the domain, so the minimum at (0,0) is an absolute (or global) minimum.
For the global maximum we have to look at the values of the function at the corners, thus
f (± 1 ,± 1 ) = 1
2 − 1 + 1
2 = 1
f (± 1 , 1 ) = 1
2
2 = 3
Thus, we got two absolute maximum points at (-‐1,1) and (1,-‐1).
Solution 2, using Lagrange multipliers:
∇
z = λ∇
( x + y − 1 )
2 x , 2 y = λ 1 , 1
x , y = λ 1 , 1
x = y
0 = x + y − 1 = 2 x − 1
x = y = 0.
or
L = x
2
2
− λ ( x + y − 1 )
Lx = 2 x − λ = 0
Ly = 2 y − λ = 0
L λ = x + y − 1 = 0
x + y − λ = 0
x + y − 1 = 0
⇒ λ = 1 ⇒ ( x , y ) = ( 0.5,0.5)
Ex 2. Find absolute min/max of f ( x , y ) = x
2
+ 2 x − 2 y on x
2
In x 2
f ' x = 2 x + 2 = 0
f ' y = 2 y − 2 = 0
and so we found an extreme value at (^) ( x , y ) = (^) ( − 1 , (^1) ) ,
however this point is outside the domain of interest.
On x
2
2 = 1 / 4
f = λ∇
g
2 x + 2 , 2 y − 2 = λ 2 x , 2 y ⇒ x + 1 , y − 1 = λ x , y
x + 1 = λ x
y − 1 = λ y
( λ − 1 ) x = 1
(^ λ −^1 ) y^ =^ −^1
⎩⎪^
⇒ x = − y
x 2
+ (− x )
2 = 2 x 2 = 1 / 4 ⇒ x = ± 1 / 8
Therefore the extreme values at (^) ( x , y ) = (^) ( 1 / 8 ,− 1 / (^8) ), (^) (− 1 / 8 , 1 / (^8) )
f ( x ,− x ) = x
2
2
2
1 / 4 + 2 / 2 max
1 / 4 − 2 / 2 min
Ex 3. Let a,b,c be lengths of edges of a triangle. Maximize the function
F = a
2
2
2
under constraint g ( b , c , θ ) =
bc sin θ = S.
Solution: Since a
2 = b
2
2 − 2 bc cos θ we rewrite
F ( a , b , c ) = a
2
2
2 = 2 b
2
2 − 2 bc cos θ = f ( b , c , θ )
Next we use Lagrange multipliers
∇
f = λ∇
g
4 b − 2 c cos θ, 4 c − 2 b cos θ, 2 bc sin θ = λ 12 c sin θ, b sin θ, bc cos θ
2 bc sin θ = λ 12 bc cos θ ⇒ λ = 4 tan θ
4 b − 2 c cos θ = 4 tan θ ⋅ 12 c sin θ
4 c − 2 b cos θ = 4 tan θ ⋅ 12 b sin θ
2 b = c ( tan θ sin θ + cos θ)
2 c = b ( tan θ sin θ + cos θ)
⎩⎪^
⇒ c = b
⇒ 2 = tan θ sin θ + cos θ =
sin 2 θ
cos θ
cos θ
a 2 = b 2
- c 2 − 2 bc cos θ = 2 b 2 − 2 b 2 cos θ = 2 b 2
( 1 − cos θ) = 2 b
2
(^1 −^12 ) =^ b
2 ⇒ a = b
Finally, we got Equilateral triangle.
