# MATH 16B Lecture Notes - Lecture 8: Minimax, Multiple Integral

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Three-variable equations

● Max/min of f(x, y, z) such that g(x, y, z) = 0

○ F(x, y, z, lambda) = f(x, y, z) + (lambda)g(x, y, z)

○ Calculate PD(x), PD(y), PD(z), PD(lambda)

○ Solve all equations when equal to 0

■ (a, b, c, d) solution --> (a, b, c) potential max/min

● Example: Determine the maximum possible volume of a cuboid such that surface area is 6.

Volume = xyz; surface area = 2xy + 2xz + 2yz = 6 (assume max volume exists and x, y, z > 0)

○ Xyz = f(x, y, z); 2xy + 2xz + 2yz = 6 --> xy + xz + yz -3 = 0 = g(x, y, z)

■ F(x, y, z, lambda) = xyz + lambda(xy) + lambda(xz) + lambda (yz) - lambda(3)

■ PD(x) = yz + lambda(y) + lambda(z), PD(y) = xz + lambda(x) + lambda(z), PD(z) =

xy + lambda(x) + lambda(y), PD(lambda) = xy + xz + yz - 3

■ Yz + lambda(y) + lambda(z) = 0, xz + lambda(x) + lambda(z) = 0, xy + lambda(x) +

lambda(y) = 0, xy + xz + yz - 3 = 0

■ Lambda = -yz/(y+z), lambda = -xz/(x + z), lambda = -xy/(x + y) -->

-yz/(y + z) = -xz/(x + z) and -xz/(x + z) = -xy/(x + y)

● -y/(y + z) = -x/(x + z) and -z/(x + z) = -y/(x + y) --> -y(x + z) = -x(y + z) and -z(x + y) = -y(x + z) --> -yz

= -xz and -xz = -xy -->

-yz = -xz and -xz = -xy --> y = x and y = z

● Xy + xz + yz - 3 = 0 --> x^2 + x^2 + x^2 - 3 = 0 --> 3x^2 - 3 = 0 --> x^2 = 1 --> x = 1

○ Y = 1 and z = 1 and lambda = -1/2

○ (1, 1, 1, -1/2) is the only solution (x, y, z > 0)

○ Max must be at (1, 1, 1) so the max volume is 1

Double Integrals

● f(x): single-variable function, continuous on [a, b]

○ Definite integral of f(x) = area above x-axis bounded by y = f(x) between a and b minus

area below x-axis bounded by y = f(x) between a and b

○ Fundamental Theorem: definite integral of f(x) = F(b) - F(a)

■ F'(x) = f(x)

● Aim: Generalize to 2-variable functions

○ Replace [a, b] with rectangle R in xy-plane

○ Replace f(x) with f(x, y) continuous 2-variable function on R

■ Definition: Double integral of f(x, y) on R = volume above xy-plane bounded by z

= f(x, y) and R minus volume below xy-plane bounded by z = f(x, y) and R

● How do we calculate the double integral of f(x, y) on R?

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