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Section 7.4: Lagrange Multipliers and

Constrained Optimization

A constrained optimization problem is a problem of the form maximize (or minimize) the functionF(x,y)subject to the conditiong(x,y) = 0.1From two to one In some cases one can solve foryas a function ofxand then find the extrema of a one variable function. That is, if the equationg(x,y) = 0 is equivalent toy=h(x), then we may setf(x) =F(x,h(x)) and then find the valuesx=afor whichfachieves an extremum. The extrema ofFare at (a,h(a)).2

Example

Find the extrema ofF(x,y) =x2y-ln(x) subject to

0 =g(x,y) := 8x+ 3y.3Solution

We solvey=-83x. Setf(x) =F(x,-83x) =-83x3-ln(x).

Differentiating we havef?(x) =-8x2-1x. Settingf?(x) = 0, we must solvex3=-18, orx=-12. Differentiating again, f ??(x) =-16x+1x2so thatf??(-12) = 12>0 which shows that-12 is a relative minimum offand (-12,43) is a relative minimum ofF subject tog(x,y) = 0.4

A more complicated example

Find the extrema ofF(x,y) = 2y+xsubject to

0 =g(x,y) =y2+xy-1.5Solution: Direct, but messy

Using the quadratic formula, we find

y=12(-x±?x2+ 4) Substituting the above expression foryinF(x,y) we must find the extrema of f(x) =?x2+ 4 and ?(x) =-?x2+ 46

Solution, continued

f ?(x) =x⎷x2+ 4 and ?(x) =-x⎷x2+ 4 Settingf?(x) = 0 (respectively,??(x) = 0) we findx= 0 in each case. So the potential extrema are (0,1) and (0,-1).7Solution, continued f ??(x) =4(⎷x2+ 4)3 and ??(x) =-4(⎷x2+ 4)3 Evaluating atx= 0, we see thatf??(0)>0 so that (0,1) is a relative minimum and as???(0)<0, (0,-1) is a relative maximum. (even thoughF(0,1) = 2>-2 =F(0,-1) !)8

Lagrange multipliers

IfF(x,y) is a (sufficiently smooth) function in two variables and g(x,y) is another function in two variables, and we define H(x,y,z) :=F(x,y) +zg(x,y), and (a,b) is a relative extremum of Fsubject tog(x,y) = 0, then there is some valuez=λsuch that

∂H∂x|(a,b,λ)=∂H∂y|(a,b,λ)=∂H∂z|(a,b,λ)= 0.9Example of use of Lagrange multipliers

Find the extrema of the functionF(x,y) = 2y+xsubject to the constraint 0 =g(x,y) =y2+xy-1.10

Solution

SetH(x,y,z) =F(x,y) +zg(x,y). Then

∂H∂x= 1 +zy ∂H∂y= 2 + 2zy+zx ∂H∂z=y2+xy-111Solution, continued Setting these equal to zero, we see from the third equation that y?= 0, and from the first equation thatz=-1y, so that from the second equation 0 = -xyimplying thatx= 0. From the third equation, we obtainy=±1.12

Another Example

Find the potential extrema of the function

f(x,y) =x2+ 3xy+y2-x+ 3ysubject to the constraint that

0 =g(x,y) =x2-y2+ 1.13Solution

SetF(x,y,λ) :=f(x,y) +λg(x,y). Then

∂F∂x= 2x+ 3y-1 + 2λx(1) ∂F∂y= 3x+ 2y+ 3 + 2λy(2) ∂F∂λ=x2-y2+ 1 (3)14

Solution, continued

Set these all equal to zero.

Multiplying the first line byyand the second byxwe obtain:

0 = 2xy+ 3y2-y+ 2λxy

0 = 2xy+ 3x2+ 3x+λxy

Subtracting, we have

0 = 3(x2-y2) + 3x-y15Solution, continued

As 0 =x2-y2+ 1, we conclude thaty= 1-3x. Substituting, we have

0 =x2-(1-3x)2+1 =x2-9x2+6x-1+1 =-8x2+6x=x(6-8x).

So the potential extrema are at (0,1) or (34,-14).16quotesdbs_dbs14.pdfusesText_20