Section 10.6: Optimization
Definition: Let f be a function with domain D ⊂ R2 containing the point (x0 , y0 ).
• The function has an absolute maximum at (x0 , y0 ) if f (x, y) ≤ f (x0 , y0 ) for all
(x, y) ∈ D. The number f (x0 , y0 ) is called the absolute maximum value of f on
D. Similarly, f has an absolute minimum at (x0 , y0 ) if f (x, y) ≥ f (x0 , y0 ) for all
(x, y) ∈ D. The number f (x0 , y0 ) is called the absolute minimum value of f on D.
The absolute maximum and minimum values of f are called the absolute extrema
of f .
• The function has a local maximum at (x0 , y0 ) if f (x, y) ≤ f (x0 , y0 ) for all (x, y)
in some disk centered at (x0 , y0 ). The number f (x0 , y0 ) is called a local maximum
value of f . Similarly, f (x, y) has a local minimum at (x0 , y0 ) if f (x, y) ≥ f (x0 , y0 )
for all (x, y) in some disk centered at (x0 , y0 ). The number f (x0 , y0 ) is called a local
minimum value of f . The local maximum and minimum values of f are called the
local extrema of f .
Theorem: (Partial Derivative Criteria)
If f has a local extremum at (x0 , y0 ) and f is differentiable at (x0 , y0 ), then
fx (x0 , y0 ) = fy (x0 , y0 ) = 0.
Definition: A point (x0 , y0 ) such that fx (x0 , y0 ) = fy (x0 , y0 ) = 0, or at least one of these
partial derivatives does not exist, is called a critical point of f .
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Example: Find the critical points of each function.
(a) f (x, y) = x2 + y 2 + 4x − 6y + 1
(b) f (x, y) = 2x2 + 2xy + y 2 − 2x − 2y
(c) f (x, y) = 2x3 + xy 2 + 5x2 + y 2
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Theorem: (Second Partials Test)
Suppose f is a function with continuous second-order partial derivatives and (x0 , y0 ) is a
critical point of f . Let
D = D(x0 , x0 ) = fxx (x0 , y0 )fyy (x0 , y0 ) − [fxy (x0 , y0 )]2 .
1. If D > 0 and fxx (x0 , y0 ) > 0, then f (x0 , y0 ) is a local minimum.
2. If D > 0 and fxx (x0 , y0 ) < 0, then f (x0 , y0 ) is a local maximum.
3. If D < 0, then f does not have a local extremum at (x0 , y0 ). The point (x0 , y0 ) is
called a saddle point. The surface below is the graph of f (x, y) = x2 − y 2 which has
a saddle point at (0, 0).
Note: The function D(x, y) is called the discriminant. If D(x0 , y0 ) = 0, the Second Partials
Test gives no information.
Example: Find the local extrema and saddle points of f (x, y) = x2 + y 2 − 2x.
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Example: Find the local extrema and saddle points of f (x, y) = 3x2 + 8y 3 + 12x − 12y 2 + 5.
Example: Find the local extrema and saddle points of f (x, y) = 3xy − x3 − y 3 .
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Example: Find three numbers whose sum is 60 and whose product is a maximum.
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Example: Find the maximum volume of an open rectangular box with surface area 75 m2 .
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Definition: A closed set in R2 is a set that contains all of its boundary points. A bounded
set in R2 is a set that is contained in some disk.
Theorem: (Extreme Value Theorem)
If f is continuous on a closed, bounded set D ⊂ R2 , then f attains an absolute maximum
and minimum value on D.
To find the absolute extrema of a continuous function f on a closed, bounded set D:
1. Find all extreme values in D.
2. Find all extreme values on the boundary of D.
3. The largest of these values is the absolute maximum and the smallest is the absolute
minimum.
Example: Find the absolute extrema of f (x, y) = 2+2x+2y −x2 −y 2 on the closed triangular
region T with vertices (0, 0), (0, 4), and (4, 0).
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Example: Find the absolute extrema of f (x, y) = x2 + y 2 − 6y + 3 on the disk
D = {(x, y)|x2 + y 2 ≤ 16}.
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Lagrange Multipliers
In many applied problems, a function of two variables, f (x, y), must be optimized subject
to a constraint of the form g(x, y) = c.
Theorem: (Lagrange’s Theorem)
Suppose that f and g are functions with continuous first-order partial derivatives and f has
an extremum at (x0 , y0 ) on the smooth curve g(x, y) = c. If ∇g(x0 , y0 ) 6= ~0, then there is a
number λ such that
∇f (x0 , y0 ) = λ∇g(x0 , y0 ).
The number λ is called a Lagrange multiplier.
Method of Lagrange Multipliers:
To find the extreme values of f (x, y) subject to the constraint g(x, y) = c,
1. Find all values of x, y, λ such that
∇f (x, y) = λ∇g(x, y),
g(x, y) = c.
2. Evaluate f at each point (x, y) found in Step 1. The largest of these values is the
maximum value of f and the smallest is the minimum value of f .
Example: Find the extreme values of f (x, y) = 6x + 8y on the circle x2 + y 2 = 25.
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Example: Suppose you want to enclose a rectangular plot of land. You have 1600 ft of fencing
available. What are the dimensions of the plot that will have the largest area?
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