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Constrained Optimization and Lagrange

Multipliers

MATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Fall 2007

Constrained Optimization

In the previous section we found the local or absolute extrema of a function either on the entire domain of the function or on a bounded region.

Now we will look for extrema which satisfy some side condition(s) known as constraint(s).

Geometry

Note: At the minimum distance from the origin the parabola and the circle are tangent. The normals to the parabola and the circle are parallel at the point of tangency. The gradient is always normal to the curve.

Geometry (cont.)

parabola: x^2 + 3 x + 2 − y = 0 circle: x^2 + y^2 = r 2

Gradients:

∇(x^2 + y^2 ) = λ∇(x^2 + 3 x + 2 − y) 〈 2 x, 2 y〉 = λ〈 2 x + 3 , − 1 〉

Equivalent system of equations:

2 x = λ( 2 x + 3 ) 2 y = −λ 0 = x^2 + 3 x + 2 − y

Method of Lagrange Multipliers

Problem: find the extreme values of f (x, y, z) subject to the constraint g(x, y, z) = 0.

Solution: Suppose f has an extremum at (x 0 , y 0 , z 0 ) on the surface S defined by g(x, y, z) = 0. Let C be a curve traced out by the vector-valued function r (t) = 〈x(t), y(t), z(t)〉 such that r (t 0 ) = 〈x 0 , y 0 , z 0 〉. Define h(t) = f (x(t), y(t), z(t)), then at the extremum h′(t 0 ) = 0.

Method of Lagrange Multipliers (cont.)

0 = h′(t 0 ) = fx (x(t 0 ), y(t 0 ), z(t 0 ))x′(t 0 ) + fy (x(t 0 ), y(t 0 ), z(t 0 ))y′(t 0 )

• fz (x(t 0 ), y(t 0 ), z(t 0 ))z′(t 0 ) = 〈fx (x 0 , y 0 , z 0 ), fy (x 0 , y 0 , z 0 ), fz (x 0 , y 0 , z 0 )〉 · 〈x′(t 0 ), y′(t 0 ), z′(t 0 )〉 = ∇f (x 0 , y 0 , z 0 ) · r ′(t 0 )

Thus ∇f (x 0 , y 0 , z 0 ) is orthogonal to r ′(t 0 ).

Since r (t) is arbitrary, ∇f (x 0 , y 0 , z 0 ) is orthogonal to S and hence parallel to ∇g(x 0 , y 0 , z 0 ).

∇f (x 0 , y 0 , z 0 ) = λ∇g(x 0 , y 0 , z 0 )

Result

Theorem Suppose that f (x, y, z) and g(x, y, z) are functions with continuous first partial derivatives and ∇g(x, y, z) 6 = 0 on the surface g(x, y, z) = 0. Suppose that either (^1) the minimum value of f (x, y, z) subject to the constraint g(x, y, z) = 0 occurs at (x 0 , y 0 , z 0 ); or (^2) the maximum value of f (x, y, z) subject to the constraint g(x, y, z) = 0 occurs at (x 0 , y 0 , z 0 ). Then ∇f (x 0 , y 0 , z 0 ) = λ∇g(x 0 , y 0 , z 0 ), for some constant λ (called a Lagrange multiplier ).

Equivalent Set of Equations

fx (x, y, z) = λgx (x, y, z) fy (x, y, z) = λgy (x, y, z) fz (x, y, z) = λgz (x, y, z) g(x, y, z) = 0

For functions of two variables this becomes:

fx (x, y) = λgx (x, y) fy (x, y) = λgy (x, y) g(x, y) = 0

Example

Example Find the extreme values of f (x, y) = 2 x^3 y subject to x^2 + y^2 = 4.

• 3 - 2 - 1 0 1 2 3
• 3
• 2
• 1

0

1

2

3

x

y

Example (cont.)

• 2
  • 1 0 1 2

x

• 2
  • 1

0

1

2 y

• 10
• 5

0

5

10

z

Example (cont.)

• (^20) x^2
• 2 0

2

y

• 5

0

5 z

Two Equality Constraints

Example Maximize f (x, y, z) = 3 x + y + 2 z subject to y^2 + z^2 = 1 and x + y − z = 0.

• 2
  • 1 0 1 2

x

• 2
  • 1

0

1

2 y

• 2
• 1

0

1

2

z

Homework

Read Section 12.8. Pages 1020-1022: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45