Constraints that are not Closed Curves

The Lagrange multiplier method is a means of finding the extrema of z(t) = f(x(t), y(t)) when the constraint g(x,y) = k is parametrized by r(t) = á x(t), y(t) ñ , t in [ a,b] . Thus, if z(t) is continuous on [a,b], then z(t) must attain an absolute maximum and an absolute minimum somewhere on r(t) = á x(t), y(t) ñ , t in [ a,b] , although it is possible that extrema will occur at the endpoints of a constraint.       

EXAMPLE 2    Find the point(s) on the curve y = x²-1 closest to the origin.       

Solution: If we let f( x,y) be the square of the distance from a point ( x,y) to the origin ( 0,0), then we are seeking to minimize

f( x,y) = x2+y2

subject to the constraint x²-y = 1. Moreover, any point (x,y) with x ³ 1 is more than one unit from the origin, so we need only find the absolute minimum over the section of the curve y = x²-1 for which x is in [ -1,1] .
       The gradients of f and g( x,y) = x²-y are

Ñf = á 2x,2y ñ ,        Ñg = á2x,-1 ñ

Thus, the Lagrange multiplier problem is

2x = l2x,        2y = -l,        x2-y = 1

To eliminate l, we use the second equation to replace l in the first equation,

2x = ( -2y) 2x,        x = -2xy

If x = 0, then y = -1. Thus, ( 0,-1) is the critical point corresponding to l = -1. If x ¹ 0, then 1 = -2y so that y = -1/2 and

x2-  -1 2 = 1,        x =  ±1 Ö2 ,

Thus, the critical points are ( 0,-1) , ( 1/Ö2,-1/2) and ( -1/Ö2,-1/2) . However,

f( -1/Ö2,-1/2) = f( 1/Ö2,1/2) =  3 4 ,        f( 0,1) = 1

Consequently, the points on y = x²-1 which are closest to the origin are

æ è  -1 Ö2 ,  -1 2 ö ø     and    æ è  -1 Ö2 ,  -1 2 ö ø

and these points are Ö(3/4) =  0.866 units from the origin.

   

Nearly all of the optimization problems considered earlier in an introductory calculus course are constrained optimization problems. Thus, they can often be solved using the method of Lagrange multipliers.       

EXAMPLE 3    Suppose John wants to start a kennel by building 5 identical adjacent rectangular runs out of 400 feet of fencing (see diagram below), Find the dimensions of each run that maximizes its area.

Solution: We let A denote the area of a run, and we let x,y be the dimensions of each run. Clearly, there are to be 10 sections of fence corresponding to widths x and 6 sections of fence corresponding to lengths y. Thus, we desire to maximize A = xy subject to the constraint

10x+6y = 400

Since x and y cannot be negative, we need only find absolute extrema for x in [ 0,40] . If we let g( x,y) = 10x+6y, then the gradients of A and g are

ÑA = á y,x ñ ,    Ñg = á10,6 ñ

As a result, the Lagrange multiplier problem becomes

y = 10l,    x = 6l,        10x+6y = 400

Solving for l in the first two equations yields l = y/10 and l = x/6. Thus,

y 10 =  x 6 ,        y =  10x 6 =  5x 3

Substituting into the constraint thus yields

10x+6 æ è  5x 3 ö ø = 400,        x = 20  feet

Moreover, we also have y = 5·20/3 = 100/3 = 33. 33 feet.