An example from last lecture (continued)
Find the absolute maximum and minimum values of the function
p
f (x, y ) = x 2 + y 2 − 2x + 2
on the closed half-disk
R = {(x, y ) : x 2 + y 2 ≤ 4 with y ≥ 0}.
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Recall : Procedure for finding absolute maxima/minima
Step 1
Find the values of f at all critical points in R.
Step 2
Find the maximum and minimum values of f on the boundary of R.
Step 3
The largest function value found in steps 1 and 2 is the absolute maximum
value of f on R. Similarly for absolute min.
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Back to the example:
Find the absolute maximum and minimum values of the function
p
f (x, y ) = x 2 + y 2 − 2x + 2
on the closed half-disk
R = {(x, y ) : x 2 + y 2 ≤ 4 with y ≥ 0}.
Last time, we completed step 1 : the critical point of f on R was found to
be (1, 0); the critical value was f (1, 0) = 1.
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Step 2 : Max/min on the boundary
For the function
f (x, y ) =
p
x 2 + y 2 − 2x + 2,
find the maximum and minimum value of f on the curved edge of R,
namely on the half-circle
x 2 + y 2 = 4,
A. max =
B. max =
√
√
6, min =
y ≥ 0.
√
10, min =
2
√
2
C. max = 10, min = 2
D. max = 6, min = 1
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Step 2 (ctd): Max/min on the boundary
For the function
f (x, y ) =
p
x 2 + y 2 − 2x + 2,
find the maximum and minimum value of f on the horizontal edge of R,
namely on the set
−2 ≤ x ≤ 2, y = 0.
A. max = 6, min = 1
B. max = 6, min = 2
√
C. max = 10, min = 2
√
D. max = 10, min = 1
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Step 3: Conclusion of the example
Combining the results from steps 1 and 2, possible candidates for
max/min are
√
√
f (1, 0) = 1, f (−2, 0) = 10, f (2, 0) = 2.
The largest of these values is
√
10, and the smallest is 1. Therefore,
Answer
absolute max = f (−2, 0) =
Math 105 (Section 204)
√
10,
absolute min = f (1, 0) = 1.
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Lesson plan: Constrained optimization
Statement of the problem
Given a function f (x, y ) (called the objective function), we need to find
the smallest and largest values of f subject to the constraint
g (x, y ) = 0.
Math 105 (Section 204)
Multivariable Calculus – Extremization
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Lesson plan: Constrained optimization
Statement of the problem
Given a function f (x, y ) (called the objective function), we need to find
the smallest and largest values of f subject to the constraint
g (x, y ) = 0.
The technique used to solve the problem above is called the method of
Lagrange multipliers.
Math 105 (Section 204)
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Method of Lagrange multipliers
Steps in the solution
Find the values of x, y and λ that satisfy the equations
∇f (x, y ) = λ∇g (x, y )
Math 105 (Section 204)
and
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g (x, y ) = 0.
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Method of Lagrange multipliers
Steps in the solution
Find the values of x, y and λ that satisfy the equations
∇f (x, y ) = λ∇g (x, y )
and
g (x, y ) = 0.
Among the solutions (x, y ) that you found in the previous step, select
the ones which gives largest and smallest values of the function f .
These are the maximum and minimum values of the objective
function subject to the constraint.
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