4452 Mathematical Modeling Lecture 4: Lagrange Multipliers

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1 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 4452 Mthemticl Modeling Lecture 4: Lgrnge Multipliers Lgrnge multipliers re high powered mthemticl technique to find the mximum nd minimum of multidimensionl functions when there re constrints dded to the problem. The technique is simple, nd esy to implement on computer. It is used extensively in physics, where I hve seen it used to determine the Bose-Einstien sttisticl distribution obeyed by bosons nd equilibrium distributions of gses. For function of two vribles, the constrint cn be thought of s curve in the xy-plne which the solution must lie on. A solution which does not lie on this curve hs to be discrded. Exmple Sy we wnt to find the mximum of function, z = f(x, y) = (x y /4)e x2 y 2, subject to the constrint tht g(x, y) = (x /2) 2 + (y /3) 2 =. If we plot the constrint curve g(x, y) = with the level curves of the function, f(x, y) = c, we get something tht looks like this: y x Figure : The constrint curve g(x, y) = is the dshed line in red, the level curves for the function f(x, y) re the solid curves in blck.

2 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 2 We solved for the globl mximum of the function f in Lecture 2, nd we found tht the mximum occurred t the point (.566,.566). However, we cn see tht this point does not lie on the constrint curve, nd therefore does not solve the present problem. Where do you think the mximum of f subject to the constrint g occurs? I think it will be round (.,.25), since tht is where the curve g seems to pproch the level curve with lrgest vlue (.235). Notice tht t this point the constrint curve g nd the level curve of the function f pper to be very similr, in the sense tht they hve the sme slope, or common tngent line. This is not coincidence! This is t the hert of the ide of Lgrnge multipliers. The concept of Lgrnge multipliers will extend to higher dimensions, but this low dimension exmple mkes visuliztion possible. The slope of n implicit function cn be found by tking the derivtive, s we do here for the implicitly defined constrint function: g(x, y) = k d [g(x, y) dx = k] d [g(x, y)] dx = g x (x, y) dx dx + g y(x, y) dy dx = g x (x, y) + g y (x, y) dy dx = dy dx = g x(x, y) g y (x, y), provided g y(x, y). A similr rgument cn be pplied to the level curve, f(x, y) = c, of the function we re trying to optimize. dy dx = f x(x, y) f y (x, y). We wnt these two slopes to be equl, nd we cn gurntee tht by hving f x (x, y) = λg x (x, y) nd f y (x, y) = λg y (x, y), where λ is sclr multiple nd is clled the Lgrnge multiplier. The Lgrnge multiplier contins the ide tht the two curves cn hve the sme slope t point, but hve very different curvture t tht point (Fig. 2). Figure 2: Two curves with the sme tngent line t point but different curvture t tht point

3 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 3 Using vector nottion nd the grdient, f(x, y) = (f x (x, y), f y (x, y)), we cn write the concept of Lgrnge Multipliers s Lgrnge Multipliers To find the mximum nd minimum vlues of z = f(x, y) subject to the constrint g(x, y) = k, you need to:. Find ll the vlues x, y, λ such tht f(x, y) = g(x, y), nd g(x, y) = k. 2. Evlute f t ll the points founds in step. The lrgest of these is the mximum vlue of f; the smllest is the minimum vlue of f. You should note tht the mx nd min cn occur t more thn one point. Let s finish off the exmple we hd been looking t, where we wnted to find the mximum of z = f(x, y) = (x y /4) e x2 y 2, subject to the constrint tht g(x, y) = (x /2) 2 + (y /3) 2 =. To solve nlyticlly: f(x, y) = (x y /4)e x2 y 2 f x (x, y) = 2 (2 + x 4x2 4xy)e x2 y 2 f y (x, y) = 2 (2 + y 4y2 4xy)e x2 y 2 g(x, y) = (x /2) 2 + (y /3) 2 g x (x, y) = 2x g y (x, y) = 2y 2/3 2 (2 + x 4x2 4xy)e x2 y 2 = λ(2x ) Solve: 2 (2 + y 4y2 4xy)e x2 y 2 = λ(2y 2 3 ) (x 2 )2 + (y 3 )2 = for (x, y, λ) Wow! This looks difficult (nd it is). I used Mthemtic to solve this system, nd even then I hd to get tricky (just sking it to solve the three equtions simultneously took long time nd I tired of witing). Eliminte[{D[f[x, y], x] == LM D[g[x, y], x],d[f[x, y], y] == LM D[g[x, y], y], g[x, y] == }, LM] Solve[{%}, {x, y}]; N[%] This is wht Mthemtic returned (I hve removed the smll complex prt). Point (, b) Functionl Vlue f(, b) Conclusion ( ,.33959) Minimum (.4854,.635).5 (.4789, ).48 (.7593,.2533).228 Mximum

4 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 4 Remember tht our guess ws (x, y) = (.,.25)? The lesson we should tke wy is sometimes grphicl solution is very vluble, nd esier to implement thn n nlytic method! But the best thing is to understnd wht you re doing so you cn use those grphicl tricks. The result is shown in Fig. 3, with level curves nd constrint. The blck dots mrk the mximum nd minimum of the function long the constrint. These plots of level curves (or level sets, s the text clls them) nd constrint curves cn be very enlightening, nd help you explin your results succinctly y x Figure 3: The result showing the mx nd min with the level curves nd contrint curve. Higher Dimension To mximize the function w = f(x, y, z) subject to the constrints g (x, y, z) = k nd g 2 (x, y, z) = k 2, you solve the system f(x, y, z) = λ g (x, y, z) + λ 2 g 2 (x, y, z) g (x, y, z) = k g 2 (x, y, z) = k 2 for x, y, z, λ, λ 2. You cn extend in similr fshion to even higher dimension. Inequlity Constrints If the problem to be solved hs inequlity constrints, then first find ny extrem using the techniques for unconstrined optimiztion, then use the Lgrnge multiplier technique for ny boundries tht exist. Compre ll the vlues you find to determine the mximum nd minimum. If you hve boundries tht chnge definition, for instnce squre, you my hve to use the Lgrnge multiplier technique more thn once to determine the extrem on the boundry. Exmple Find the mximum nd minimum of f(x, y, z, t) = x 2 + y 2 + z 2 + t 2 subject to the constrints x z 2 nd y 2 + t 4.

5 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 5 Since this is higher dimension problem, it my be difficult to visulize. We lso hve n inequlity restrint. The inequlity restrint mens we should first find the etrem s if there were no contrints. If the extrem lies within the constrined region, then we re done. If the extrem does not lie within the constrined region, then it must lie on the boundry. We will then proceed to use Lgrnge multipliers to find the extrem. First, the unconstrined optimiztion. For this we must solve f x (x, y, z, t) =, f y (x, y, z, t) =, f z (x, y, z, t) =, f t (x, y, z, t) =, for (x, y, z, t). This yields the solution (,,, ). This is obviously minimum, since the function f(x, y, z, t). The point (,,, ) stisfies the constrints, x z 2, nd y 2 + t 4, so this will be the minimum of the function subject to the constrints. Since we found no mximum inside the region, the mximum must occur on the boundry. We will use Lgrnge multipliers to find it. We wnt to solve the system: f x (x, y, z, t) = λ (g ) x (x, y, z, t) + λ 2 (g 2 ) x (x, y, z, t) f y (x, y, z, t) = λ (g ) y (x, y, z, t) + λ 2 (g 2 ) y (x, y, z, t) f z (x, y, z, t) = λ (g ) z (x, y, z, t) + λ 2 (g 2 ) z (x, y, z, t) f t (x, y, z, t) = λ (g ) t (x, y, z, t) + λ 2 (g 2 ) t (x, y, z, t) g (x, y, z, t) = x z = 2 g 2 (x, y, z, t) = y 2 + t = 4 for (x, y, z, t, λ, λ 2 ). Using Mthemtic to do this, we find: Lgrnge Multipliers (λ, λ 2 ) Point (x, y, z, t) Functionl Vlue f(x, y, z, t) Conclusion (2, 8) (,,, 4) 8 Mximum (2, ) (, 7/2,, /2) 5.75 (2, ) (, 7/2,, /2) 5.75 Notice tht the minimum on the boundry of the constrints occurs t two plces, nd is lrger thn the minimum we found inside the constrint region. The mximum of the function f subject to the constrints g nd g 2 is 8. Sensitivity Anlysis Sensitivity nlysis for Lgrnge Multipliers proceeds in much the sme wy s before, we tke prmeter in the problem, chnge it to vrible, nd proceed to solve. Exmple We wnt to find the sensitivity to the prmeter α of the mximum of z = f(x, y) = ( 4 x + αy ) ( x 45 5x ) y + 3z,

6 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 6 subject to the constrint tht g(x, y) = 4x + y + z =, You cn verify tht for α = 6, the mximum is which occurs t the point (49.364, 288.6, ). Solving this vi the Lgrnge multiplier method, we find f x (x, y, z) = 4 x 5 + y 2 + αy f y (x, y, z) = 45 + x 2 + αx f z (x, y, z) = 3 g x (x, y, z) = 4 g y (x, y, z) = g z (x, y, z) = Solve: 4 x 5 + y 2 + αy = 4λ 45 + x 2 + αx = λ 3 = λ 4x + y + z = for (x, y, z, λ) Lgrnge Multiplier λ ( Point (x, y, z) ( ), 5 + α (5 + ) 2, 2 ( ) ) (5 + ) 2 If we desire S(x, α), we cn get it by differentiting: Similrly, S(x, α) = dx dα α x = α 5 + α. S(y, α) = dy dα α 2 α ( α) = y (5 + α) ( α) S(z, α) = dz dα α z = 2 α ( α) (5 + α) ( α + 5 α 2 ) S(f, α) = df dα α f = 32 α ( α) (5 + α) ( α + α 2 ) The grphs of the sensitivities re contined in Fig. 4. Since we re fr wy (α 6) from ll the res where the sensitivity is lrge, we cn sfely sy tht the mximum is robust with respect to the prmeter α being pproximtely 6.

7 Mth Modeling Lecture 4: Lgrnge Multipliers Pge 7 Sensitivity of Optiml x Sensitivity of Optiml y Sensitivity of Optiml z Sensitivity of Mximum f Figure 4: The sensitivity grphs.

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