11.8 Constraint Optimization: Lagrange's Multipliers

Arkansas Tech University
MATH 2934: Calculus III

Dr. Marcel B Finan
11.8 Constraint Optimization: Lagranges Multipliers

Most optimization problems encountered in the real world are constrained by
external circumstances. For example, a city wanting to improve its roads has
only limited number of tax dollars it can spend on the project.
Constrained optimization is the maximization or minimization of an objec-
tive function subject to constraints on the possible values of the independent
variable(s). Constraints can be either equality constraints or inequality con-
straints. In this section, we see how to find an optimum value of a function
of two variables subject to some constraints using a graphical approach and an
analytical approach that employs the so-called Lagranges Multipliers.
Graphical Approach
We consider the following example. A company has determined that its pro-
2 1
duction function (in units) is the Cobb-Douglas function f (x, y) = x 3 y 3 where
x and y are raw materials. If x and y cost $1000 per unit and $3780 is the
budget available , then what is the maximum production that can be obtained
that exhausts the available budget?
Mathematically, we are asked to maximize the function f subject to the con-
straint g(x, y) = 3.78 where g(x, y) = x + y. Graphically, the line x + y = 3.78
represents all the combinations of raw materials that just exhaust the budget
but are still affordable.
The maximum production can be located graphically by plotting contours of f
on a plot containing the line x + y = 3.78 as shown in Figure 11.8.1.
To maximize f we find the point which lies on the level curve with the largest
possible value of f and which lies on the line x + y = 3.78. This figure shows
that at the maximum, f must be tangent to the constraint line. Note that the
maximum value of f is about 2 and this occurs at x 2.52 and y 1.26.
In general, the global maximum or global minimum occurs where the graph
of the constraint equation is tangent to one of the level curves of the original
function.
Figure 11.8.1
1
Analytical Approach: Lagrange Multipliers
As noted above, the maximum production is achieved at the point where the
constraint is tangent to a level curve of the production function. The method of
Lagrange Multipliers uses this fact in algebraic form to calculate the maximum.
From Figure 11.8.1, we see that at the point of tangency, f (x, y) and g(x, y)
are parallel so that f (x, y) = g(x, y) for some which we call the Lagrange
multiplier. We therefore have
s
r 2
2 3 y~ 1 3 x ~
i+ j = ~i + ~j.
3 x 3 y
Hence,
r
2
2 3 y 1 3 x
p
3 x = and 3 y = .
Eliminating gives
r
2
2 3 y 1 x
p 3
3 x = 3 y which leads to 2y = x.
Substituting this into the equation x + y = 3.78 we find x = 2.52 and y = 1.26.
Hence,
2 1
f (2.52, 1.26) = (2.52) 3 (1.26) 3 2.
As before, we see that the maximum value of f is approximately 2. Also, note
that 0.53
Generalization
We are given a constraint equation
g(x, y) = c
and an objective function f (x, y). The goal is to find the global maximum and
minimum values of f among the values taken on by f along the constraint curve;
i.e., the set of points for which g(x, y) = c. Moreover, we would like to find all
of the points (x, y) at which these maxima and minima are attained. This is
provided by the following theorem.
Theorem 11.8.1
Suppose that g(x, y) 6= ~0 on the curve g(x, y) = c. If there is a maximum or
a minimum of the values that the function f (x, y) assumes on the constraint
curve g(x, y) = c, then it occurs at a point (x, y) at which
f (x, y) = g(x, y) and g(x, y) = c.

Compute the values of f at these points. The largest value is the global maxi-
mum and the smallest value is the global minimum.
2
Example 11.8.1
Find the maximum and minimum of f (x, y) = 5x 3y subject to the constraint
x2 + y 2 = 136.
Solution.
Since f (x, y) = 5~i 3~j and g(x, y) = 2x~i + 2y~j, we must have 2x = 5
5 3
and 2y = 3. Eliminating we find 2x = 2y and this leads to y = 53 x.
9 2
Substituting into the constraint equation we find x2 + 25 x = 136. Solving
this equation for x we find x = 10. If x = 10 then y = 6 and if x = 10
then y = 6. Note that g 6= ~0 on the circle. Since, f (10, 6) = 68 and
f (10, 6) = 68, the maximum of f occurs at the point (10, 6) and the mini-
mum occurs at the point (10, 6)
Optimization with Closed and Bounded Constraints

If the constraint is a closed and bounded region then by the Extreme Value
Theorem of the previous section, the global maximum or minimum occurs ei-
ther at the critical points of f inside the region or at a point on the boundary
of the region. Lets work an example to see how these kinds of problems work.
Example 11.8.2
Find the maximum and minimum values of f (x, y) = 4x2 + 10y 2 on the disk
x2 + y 2 4.
Solution.
The first step is to find all the critical points that are inside the disk (i.e., satisfy
the constraint x2 + y 2 < 4.). We have
fx (x, y) =8x = 0
fy (x, y) =20y = 0
So (0, 0) is the only critical point of f satisfying x2 + y 2 < 4.

Next, we proceed with Lagrange Multipliers and we treat the constraint as an
equality instead of the inequality. Since f (x, y) = 8x~i + 20y~j and g(x, y) =
2x~i + 2y~j, we must have 2x = 8x and 2y = 20y. These equations imply
2x( 4) =0
2y( 10) =0.
If x = 0 we find y = 2. If x 6= 0 then = 4 and so the second equation gives

y = 0 and so x = 2. If we had performed a similar analysis on the second
equation we would arrive at the same points.
So, Lagrange Multipliers gives us four boundary points to check : (0, 2), (0, 2), (2, 0), (2, 0).
Now, since
f (0, 0) =0
f (2, 0) =f (2, 0) = 16
f (0, 2) =f (0, 2) = 40
3
the global maximum of f occur at the points (0, 2) and (0, 2) and the global
minimum occurs at (0, 0)
Practical Interpretation of
Let (x , y ) be an optimum value. Then its location depends on c where
g(x, y) = c. Thus, x = x (c) and y = y (c). Moreover, we can look to f
as a composite function f (x (c), y (c)). Using the chain rule we can write
df f dx f dy
= + .
dc x dc y dc
However, we have
f g f g
x (x , y ) = x (x , y ) and
y (x , y ) = y (x , y )
so that
df g dx g dy dg
= + = .
dc x dc y dc dc
dg
Since g(x (c), y (c)) = c we must have dc = 1. Thus,
df
= .
dc
This says that is the rate of change of the optimum value of f as c increases.
For example, in the budget function discussed earlier, an increase of $1000 in
the budget will lead to an increase of about 0.53 unit in production.

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Arkansas Tech University

MATH 2934: Calculus III

11.8 Constraint Optimization: Lagranges Multipliers

f (x, y) = g(x, y) and g(x, y) = c.

Optimization with Closed and Bounded Constraints

So (0, 0) is the only critical point of f satisfying x2 + y 2 < 4.

If x = 0 we find y = 2. If x 6= 0 then = 4 and so the second equation gives

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