1. Maximum and Minimum Values

2. Maximum and Minimum Values
Problem: Find extrema (maximum or minimum) of 𝑓(𝑥,𝑦) on some domain D.
Consider a function 𝑧=𝑓 𝑥,𝑦 .
We say that (𝑎,𝑏)is a
Local maximum if 𝑓(𝑎,𝑏)≥𝑓(𝑥,𝑦) for (𝑥,𝑦) in D “around” (𝑎,𝑏)
Absolute maximum if 𝑓(𝑎,𝑏)≥𝑓(𝑥,𝑦)
for all (𝑥,𝑦) in D
Local minimum if 𝑓(𝑎,𝑏)≤𝑓(𝑥,𝑦) for (𝑥,𝑦) in D “around” (𝑎,𝑏)
Absolute minimum if 𝑓(𝑎,𝑏)≤𝑓(𝑥,𝑦) for all (𝑥,𝑦) in D

3. Maximum and Minimum Values
Theorem: If 𝑓 𝑥,𝑦 has a local maximum or minimum at a point 𝑎,𝑏 , and the first-order partial derivatives of 𝑓 exist there, then
𝑓 𝑥 𝑎,𝑏 =0 and 𝑓 𝑦 𝑎,𝑏 =0.
Geometrically:
If the function f has a local maximum (or minimum) at 𝑃(𝑎,𝑏), then the tangent plane must be horizontal with equation 𝑧=𝑓(𝑎,𝑏).
Since the equation of the tangent plane is
𝑧=𝑓 𝑎,𝑏 + 𝑓 𝑥 𝑎,𝑏 𝑥−𝑎 + 𝑓 𝑦 (𝑎,𝑏)(𝑦−𝑏)
it follows that 𝑓 𝑥 𝑎,𝑏 = 𝑓 𝑦 𝑎,𝑏 =0

4. Maximum and Minimum Values
NOTES:
is a necessary but not sufficient condition for a maximum or a minimum. We could have a saddle point at (a,b)
or or both may not exist at a local maximum or minimum.
A maximum or minimum could be on the boundary of the domain D.

5. Maximum and Minimum Values
Let 𝑎,𝑏 be a point in the interior of the domain D.
We say (𝑎,𝑏) is a critical point for 𝑓(𝑥,𝑦) if either
𝑓 𝑥 𝑎,𝑏 =0 and 𝑓 𝑦 𝑎,𝑏 =0 , or
𝑓 𝑥 or 𝑓 𝑦 or both do NOT exist at (𝑎,𝑏)
Critical points are candidates for maximum and minimum.
Example 1: Find the critical points of 𝑓 𝑥,𝑦 = 𝑥 2 + 𝑦 2 −6𝑥−4𝑦+2
𝑓 𝑥 𝑥,𝑦 =2𝑥− 𝑓 𝑦 𝑥,𝑦 =2𝑦−4
The partial derivatives are equal to 0, when 𝑥=3
and 𝑦=2, so the only critical point is 3,2 .
From the figure we can see that 𝑓 3,2 =−11 is
a minimum. The surface is the elliptic paraboloid with
vertex at 3, 2, −11 .

6. Maximum and Minimum Values
How do we determine analytically whether a critical point is a maximum, a minimum or neither?
Second Derivative Test:
We define the Discriminant:
Assume the second partial derivatives of 𝑓 are continuous “around” (𝑎,𝑏) and 𝑓 𝑥 𝑎,𝑏 = 𝑓 𝑦 𝑎,𝑏 =0
𝑫(𝒂,𝒃)
𝒇 𝒙𝒙 (𝒂,𝒃)
Classification of (𝒂,𝒃) +
Local Minimum −
Local Maximum
Saddle point
Test is inconclusive

7. Maximum and Minimum Values – Example 2
Find and classify the critical points of 𝑓 𝑥,𝑦 =4+ 𝑥 3 + 𝑦 3 −3𝑥𝑦
Contour plot
𝑓 𝑥 =3 𝑥 2 −3𝑦 𝑓 𝑦 =3 𝑦 2 −3𝑥
Setting the partial derivatives equal to 0, yields
𝑥 2 −𝑦=0 and 𝑦 2 −𝑥=0
Substitute 𝑦= 𝑥 2 from the first equation into
the second equation. This gives
0= 𝑥 4 −𝑥=𝑥( 𝑥 3 −1) with roots 𝑥=0 and 𝑥=1.
The two critical points are (0,0) and (1,1)
Critical point
𝒇 𝒙𝒙
𝒇 𝒚𝒚
𝒇 𝒙𝒚
Discriminant
𝑫= 𝒇 𝒙𝒙 𝒇 𝒚𝒚 − 𝒇 𝒙𝒚 𝟐
Type
(0,0) −3
0 0 − −3 2 =−9
Saddle since 𝐷<0
(1,1) 6
6 6 − −3 2 =27
Local minimum since 𝐷>0 and 𝑓 𝑥𝑥 >0

8. Maximum and Minimum Values - Application
Find the point (x, y, z) on the plane z = 4x + 3y + 3 which is closest to the origin.
We will minimize the square of the distance from the point (x, y, z) to the origin, 𝑑 2 = 𝑥 2 + 𝑦 2 + 𝑧 2 , subject to the constraint z = 4x + 3y + 3
Substituting z into 𝑑 2 yields
𝑓 𝑥,𝑦 = 𝑥 2 + 𝑦 2 + (4𝑥+3𝑦+3) 2
Find the critical points:
𝑓 𝑥 =2𝑥+8 4𝑥+3𝑦+3 =34𝑥+24𝑦+24=0
𝑓 𝑦 =2𝑦+6 4𝑥+3𝑦+3 =24𝑥+20𝑦+18=0
Solving the system gives
Substituting into the equation of the plane gives
The point on the plane closest to the origin is

9. Maximum and Minimum Values
ABSOLUTE MAXIMUM AND MINIMUM VALUES
EXTREME VALUE THEOREM: If 𝑓 is continuous on a closed, bounded region D in R2, then f attains an absolute maximum 𝑓( 𝑥 1 , 𝑦 1 ) and an absolute minimum 𝑓( 𝑥 2 , 𝑦 2 ) at some points ( 𝑥 1 , 𝑦 1 ) and ( 𝑥 2 , 𝑦 2 ) in D.
To find absolute maximum and minimum:
Step 1. Find the values of f at the critical points
Step 2. Find the extreme value of f on the boundary of D
Step 3. The largest value in 1. and 2. is the absolute maximum, the
smallest value is the absolute minimum.

10. Maximum and Minimum Values - Example 3
Find the absolute maximum and minimum of the function
𝑓 𝑥,𝑦 =𝑥𝑦−5𝑦−25𝑥+125
on the region D bounded by y = x2 and y = 36.
Step1: Find the critical points:
𝑓 𝑥 =𝑦−25=0, 𝑓 𝑦 =𝑥−5= (5,25)
Evaluate the function: f(5, 25) = 0
Step 2: Find the extreme values of 𝑓on the boundary of D
The boundary consists of the line segment 𝑦=36 and the parabola.
We first check the line segment.
Substituting y = 36 into the expression for 𝑓(𝑥,𝑦) gives:
𝑓 𝑥,36 =36𝑥−5 36 −25𝑥+125=11𝑥− −6≤𝑥≤6
This is an increasing function of x, so its minimum value is
𝑓 −6,36 =−121 and its maximum value is 𝑓 6,36 =11.

11. Maximum and Minimum Values - Example 3 continued
Now let’s check the boundary 𝑦= 𝑥 2 . Substituting 𝑦= 𝑥 2 into the expression for 𝑓(𝑥,𝑦), gives
𝑓 𝑥, 𝑥 2 =𝑔 𝑥 = 𝑥 3 −5 𝑥 2 −25𝑥 −6≤𝑥≤6
We can see from the graph of 𝑔 𝑥 =𝑓(𝑥 ,𝑥 2 ) that the function attains its minimum at 𝑥=−6.
To find the maximum we solve 𝑔 ′ 𝑥 =3 𝑥 2 −10𝑥−25=0. The roots are 𝑥=− 5 3 and 𝑥=5.
Values of 𝑓:
Step 3: Compare all the values from Step 1 and 2.
𝑓 5,25 =0
𝑓 −6,36 =− Absolute minimum
𝑓 −6,36 =−121
𝑓 6,36 =11
𝑓 − 5 3 , ≈ Absolute maximum
𝑓 − 5 3 , ≈148.15