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4.8 Lagrange multipliers

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Method of lagrange multipliers: one constraint

Let f and g be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve g ( x , y ) = 0 . Suppose that f , when restricted to points on the curve g ( x , y ) = 0 , has a local extremum at the point ( x 0 , y 0 ) and that ∇ g ( x 0 , y 0 ) ≠ 0 . Then there is a number λ called a Lagrange multiplier    , for which

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

Proof

Assume that a constrained extremum occurs at the point ( x 0 , y 0 ) . Furthermore, we assume that the equation g ( x , y ) = 0 can be smoothly parameterized as

x = x ( s ) and y = y ( s )

where s is an arc length parameter with reference point ( x 0 , y 0 ) at s = 0 . Therefore, the quantity z = f ( x ( s ) , y ( s ) ) has a relative maximum or relative minimum at s = 0 , and this implies that d z d s = 0 at that point. From the chain rule,

d z d s = ∂ f ∂ x · ∂ x ∂ s + ∂ f ∂ y · ∂ y ∂ s = ( ∂ f ∂ x i + j ∂ x ∂ s ) + ( ∂ f ∂ y · ∂ y ∂ s ) = 0 ,

where the derivatives are all evaluated at s = 0 . However, the first factor in the dot product is the gradient of f , and the second factor is the unit tangent vector T ( 0 ) to the constraint curve. Since the point ( x 0 , y 0 ) corresponds to s = 0 , it follows from this equation that

∇ f ( x 0 , y 0 ) · T ( 0 ) = 0 ,

which implies that the gradient is either 0 or is normal to the constraint curve at a constrained relative extremum. However, the constraint curve g ( x , y ) = 0 is a level curve for the function g ( x , y ) so that if ∇ g ( x 0 , y 0 ) ≠ 0 then ∇ g ( x 0 , y 0 ) is normal to this curve at ( x 0 , y 0 ) It follows, then, that there is some scalar λ such that

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

□

To apply [link] to an optimization problem similar to that for the golf ball manufacturer, we need a problem-solving strategy.

Problem-solving strategy: steps for using lagrange multipliers

  1. Determine the objective function f ( x , y ) and the constraint function g ( x , y ) . Does the optimization problem involve maximizing or minimizing the objective function?
  2. Set up a system of equations using the following template:
    ∇ f ( x 0 , y 0 ) = λ ∇ g ( x 0 , y 0 ) g ( x 0 , y 0 ) = 0 .
  3. Solve for x 0 and y 0 .
  4. The largest of the values of f at the solutions found in step 3 maximizes f ; the smallest of those values minimizes f .

Using lagrange multipliers

Use the method of Lagrange multipliers to find the minimum value of f ( x , y ) = x 2 + 4 y 2 − 2 x + 8 y subject to the constraint x + 2 y = 7 .

Let’s follow the problem-solving strategy:

  1. The optimization function is f ( x , y ) = x 2 + 4 y 2 − 2 x + 8 y . To determine the constraint function, we must first subtract 7 from both sides of the constraint. This gives x + 2 y − 7 = 0 . The constraint function is equal to the left-hand side, so g ( x , y ) = x + 2 y − 7 . The problem asks us to solve for the minimum value of f , subject to the constraint (see the following graph).
    Two rotated ellipses, one within the other. On the largest ellipse, which is marked f(x, y) = 26, there is a tangent line marked with equation x + 2y = 7 that appears to touch the ellipse near (5, 1).
    Graph of level curves of the function f ( x , y ) = x 2 + 4 y 2 − 2 x + 8 y corresponding to c = 10 and 26 . The red graph is the constraint function.
  2. We then must calculate the gradients of both f and g :
    ∇ f ( x , y ) = ( 2 x − 2 ) i + ( 8 y + 8 ) j ∇ g ( x , y ) = i + 2 j .

    The equation ∇ f ( x 0 , y 0 ) = λ ∇ g ( x 0 , y 0 ) becomes
    ( 2 x 0 − 2 ) i + ( 8 y 0 + 8 ) j = λ ( i + 2 j ) ,

    which can be rewritten as
    ( 2 x 0 − 2 ) i + ( 8 y 0 + 8 ) j = λ i + λ j .

    Next, we set the coefficients of i and j equal to each other:
    2 x 0 − 2 = λ 8 y 0 + 8 = 2 λ .

    The equation g ( x 0 , y 0 ) = 0 becomes x 0 + 2 y 0 − 7 = 0 . Therefore, the system of equations that needs to be solved is
    2 x 0 − 2 = λ 8 y 0 + 8 = 2 λ x 0 + 2 y 0 − 7 = 0.
  3. This is a linear system of three equations in three variables. We start by solving the second equation for λ and substituting it into the first equation. This gives λ = 4 y 0 + 4 , so substituting this into the first equation gives
    2 x 0 − 2 = 4 y 0 + 4 .

    Solving this equation for x 0 gives x 0 = 2 y 0 + 3 . We then substitute this into the third equation:
    ( 2 y 0 + 3 ) + 2 y 0 − 7 = 0 4 y 0 − 4 = 0 y 0 = 1.

    Since x 0 = 2 y 0 + 3 , this gives x 0 = 5 .
  4. Next, we substitute ( 5 , 1 ) into f ( x , y ) = x 2 + 4 y 2 − 2 x + 8 y , gives f ( 5 , 1 ) = 5 2 + 4 ( 1 ) 2 − 2 ( 5 ) + 8 ( 1 ) = 27 . To ensure this corresponds to a minimum value on the constraint function, let’s try some other values, such as the intercepts of g ( x , y ) = 0 , Which are ( 7 , 0 ) and ( 0 , 3.5 ) . We get f ( 7 , 0 ) = 35 and f ( 0 , 3.5 ) = 77 , so it appears f has a minimum at ( 5 , 1 ) .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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