Lecture 18 - Wednesday, Feb 18
MORE ON OPTIMIZATION (§14.7)
Example (Midterm II, Lieblich, Spring 2013, Ex 1)
For f(x, y) = sin(x) · cos(y)
a) Calulate fx (a, b) and fy(a, b)
b) Find all critical points of f and classify them according to their type (max, min, saddle point)
For a),
fx (a, b) = cos(a) · cos(b)
fy(a, b) = − sin(a) · sin(b)
Let us first solve b) only for 0 ≤ a, b < π.
fx (a, b) = 0 ⇔ cos(a) · cos(b) = 0 ⇔ cos(a) = 0 or cos(b) = 0
fy(a, b) = 0 ⇔ − sin(a) · sin(b) = 0 ⇔ sin(a) = 0 or sin(b) = 0
Observe: There is no a ∈ R where sin(a) = 0 = cos(a).
Only options where both derivatives are 0 simultaneuosly
are
(cos(a) = 0 and sin(b) = 0) or (sin(a) = 0 and cos(b) = 0)
π
π
⇔ (a, b) = ( , 0) or (a, b) = (0, )
2
2
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Example (cont). The 2nd derivatives are
fxx = − sin(x) cos(y)
fxy = − cos(x) sin(y)
− sin( π ) cos(0)
π
2
D( , 0) = − cos( π ) sin(0)
2
2
−sin(0) cos( π )
π
2
D(0, ) = π
− cos(0) sin( )
2
2
fyy = − sin(x) cos(y)
−1
=
0
π − cos(0) sin( 2 ) 0
=
π −1
− sin(0) cos( 2 )
− cos( π2 ) sin(0)
− sin( π2 ) cos(0)
⇒ ( π2 , 0) is local maximum, (0, π2 ) is saddle point.
0 =1
−1 −1 = −1
0 Observation:
f is periodic, i.e. f(x + 2π, y) = f(x, y) = f(x, y + 2π).
Moreover f(x + π, y) = f(x, y + π) = −f(x, y).
• {( π2 + π · s, π · t) : s, t ∈ Z, s + t even} are local maxima
• {( π2 + π · s, π · t) : s, t ∈ Z, s + t odd} are local minima
• {(π · s, π2 + π · t) : s, t ∈ Z} are saddle points
2π
π
0
0
π
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2π
Example (Midterm II, Aut. 2012, Bekyel)
Find the points on the cone given by z 2 = x 2 + y2
that are closest to the point (4, 2, 0).
Goal: Minimize h(x, y, z) = (x −4)2 +(y−2)2 +z 2 over cone.
Idea: Better optimize
f(x, y) = h(x, y, x 2 + y2) = (x − 4)2 + (y − 2)2 + x 2 + y2
= 2x 2 − 8x + 2y2 − 4y + 20
over (x, y) ∈ R2. Take partial derivatives
!
fx (x, y) = 4x − 8 = 0 ⇒ x = 2
!
fy(x, y) = 4y − 4 = 0 ⇒ y = 1
Hence (x, y) = (2, 1) is a critical point.
fxx = 4,
fxy = 0,
fyy = 4
Then
4 0
D(2, 1) = = 16 > 0
0 4
√
√
hence (2, 1) is minimum for f. Hence (2, 1, 5) and (2, 1, − 5)
are points on cone that are closest to (4, 2, 0).
Remark:
• Instead of minimizing
p
f(x, y), better minimize f(x, y)
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Example (Midterm II, Spring 2011, Loveless,1b)
Find the absolute maximum and minimum value of
f(x, y) = xy2 + x + 2 over the region
R = {(x, y) : y ≥ 0, x 2 + y2 ≤ 8}.
• Step 1: Draw the region R
y√
8
√ x
8
√
− 8
• Step 2: Find critical points in R.
fx = y2 + 1,
fy = 2xy
But y2 + 1 > 0 ⇒ no critical point in the interior of R.
• Step 3: Optimize
√
√
g(x) = f(x, 0) = x + 2 for − 8 ≤ x ≤ 8.
Since g′(x) = 2, the only candidate extrema are the
√
√
√
interval boundaries x = ± 8 (i.e. (− 8, 0) and ( 8, 0)).
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Example (cont)
• Step 4: Optimize f(x, y) over x 2 + y2 = 8. Equivalently,
optimize h(x) = x(8 − x 2) + x + 2 = −x 3 + 9x + 2.
√
′
2
2
h (x) = −3x + 9 = 0 ⇒ x = 3 ⇒ x = ± 3.
√ √
√ √
Candidate points ( 3, 5) and (− 3, 5)
• Step 5: Compare candidates
point
√
(− 8, 0)
√
( 8, 0)
√ √
(− 3, 5)
√ √
( 3, 5)
Thus
value
√
2 − 2 2 ≈ −0.828
√
2 + 2 2 ≈ 4.828
√
2 − 6 3 ≈ −8.392
√
2 + 6 3 ≈ 12.392
√
abs. maximum = 2 + 6 3,
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√
abs. mininum = 2 − 6 3
Optimization Summary:
Task: Find and classify critical points of f(x, y).
(1) Find critical points (a, b) with fx (a, b) = 0 = fy(a, b)
(2) Apply 2nd derivative test



< 0 ⇒ saddle point
f (a, b) f (a, b) 


xy
xx
< 0 local max
D(a, b) = fxy(a, b) fyy(a, b) 

> 0 ⇒ fxx (a, b)


> 0 local min

Task: Maximize f(x, y, z) on some 2D surface z = z(x, y)
(1) Eliminate one variable and write g(x, y) := f(x, y, z(x, y))
(2) Compute gx , gy, find critical points
(3) Classify using 2nd derivative test
Task: Maximize f(x, y) on a region R ⊆ R2
(1) Find critical points of f over R2
(2) Write boundary of R as 1-dim function, say g(x).
(3) Find critical points of g
WARNING: one might need to split boundary into several segments
(4) Compare candidates
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