MATH 16B Lecture Notes - Lecture 5: Partial Derivative, Saddle Point, Minimax

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21 Apr 2017
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Department
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Max / Min
Definition: We say f has a relative max at (a, b) if f(a, b) >= f(x, y) for all(x, y) sufficiently near (a,
b)
Will appear as a hill
Definition: We say f has a relative min at (a, b) if f(a, b) <= f(x, y) for all(x, y) sufficiently near (a,
b)
Will appear as a valley
For single-variable versions:
Find the horizontal line tangent to the curve: f'(a) = 0
For 2-variable versions
flat tangent plane is parallel to xy-plane
First derivative test: f has a relative max/min at (a, b)
Partial derivative (x) (a, b) = 0 and PD (y) (a, b) = 0
OBSERVATION: If PD (x) (a, b) = 0 = PD (y) (a, b), we say (a, b) is a critical point of f
Example: f(x, y) = 2 - x^2 - y^2 => PD (x) = -2x, PD (y) = -2y
X and Y both have to equal 0 so (0, 0) is the only critical point
Converse of 1st Derivative Test is false
Example: a saddle point
3-D graph that looks like a saddle; the point in the middle (a, b) is a critical point
because there can be a "flat" tangent plane there
Finding critical points:
Single-variable
f'(a) = 0 (critical points)
f''(a) > 0 => relative min
f''(a) < 0 => relative max
f''(a) = 0 => inconclusive
2-variable
D(x, y) = discriminant of f
D(x, y) = 2nd PD(x) * 2nd PD(y) - (PD(x, y))^2
2nd derivative test: suppose (a, b) is a critical point (PD (x) (a, b) = 0 and PD (y) (a, b) = 0)
D(a, b) > 0 and 2nd PD (x) (a, b) > 0 => relative min
D(a, b) > 0 and 2nd PD (x) (a, b) < 0 => relative max
D(a, b) < 0 => saddle point
D(a, b) = 0 => inconclusive
Strategy to find max/mins
Calculate PD (x) and PD (y), then solve PD (x) = 0 and PD (y) = 0. This gives critical points
Calculate 2nd PD (x) and D(x, y). Evaluate at critical points and draw conclusions from 2nd
derivative test
Example: f(x, y) = x^3 - 12x + 6y =>
PD (x) = 3x^2 - 12
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