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

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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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