LAGRANGE MULTIPLIERS

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Jesse Hill
6 years ago
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1 LAGRANGE MULTIPLIERS In our above variational methods course we briefl discussed Lagrange Multipliers and showed how these ma be used to find the etremum of a function F subject to a set of constraints. We want to here discuss this procedure in more detail and work out several more specific eamples of possible interest to the readers. Consider a function of n variables given as- F,, 3, n plus constraints g,, 3, n c, g,, 3, n c,..etc where the c n s are constants. Geometricall one can think of g n c n as hpersurfaces which intersect in a common curve C. The gradients of the various g surfaces will be at right angles to the intersection curve as shown- Here g g and g h in order to simplif the discussion. Note net that the directional derivative of the function F to be etremized and hence have df/ds is- df ds F τ whereτ is the unit length tan gent vector along curve C Thus we have that the gradient of F is also perpendicular to curve C and hence that gradg, gradh and gradf are coplanar. This means in general that- [ F g g g...] 3 3 where the s are the Lagrange Mulipliers. These n equations plus the equations for the constraints constitute sufficient information to find all s plus determine the etremum value for F.

Lets consider a few eamples starting with the simple problem of determining the maimum volume V contained in a clinder of radius R and height H for fied surface area S.

2 Lets consider a few eamples starting with the simple problem of determining the maimum volume V contained in a clinder of radius R and height H for fied surface area S. Here one has- [ π R H πr πrh ] Thus one finds the following three equations- RH R H, R, S πr R H with solutions - / πr R, H R, S 6 The result states that for maimum possible storage a can should have its diameter just equal to its height. There is an interesting eperiment carried our b child pschologists in which the fill both a tall and a wide drinking glass full of the same volume of fruit juice and then ask a child which glass contains the larger amount of fluid. Invariabl the child will choose the taller glass. Directl related to this maimum volume problem of a clinder is the Archimedes observation that the maimum volume sphere which can be put in a clinder requires that the clinder height just equals its diameter. Under those conditions one has that the sphere to clinder volume is eactl /3. A graph of this geometrand one related to that supposedl engraved on Archimedes s tombstone is- Net let us ask what is the radius of the largest volume rectangular solid which can fit into a unit radiusr sphere. Here the Lagrange Multiplier method produces-

[ z z or the equivalent- z, z, and z These ield the solutions z/sqrt3. That is, the largest volume rectangular solid capable of fitting into a unit radius sphere is a cube with sides of length /sqrt3.

3 [ z z or the equivalent- z, z, and z These ield the solutions z/sqrt3. That is, the largest volume rectangular solid capable of fitting into a unit radius sphere is a cube with sides of length /sqrt3. We show ou here a 3D picture of this result constructed via MAPLE- Note that the ratio of cube volume V c 8 3 to that of the sphere volume Vs4π/3 is /[πsqrt3] As a third, more difficult, eample, consider the shortest distance from the parabola and the circle -. This time one wants to etremize the distance squared between the two curves subjected to the curve constraints. One has the following picture-

4 We find- ] [ which produces the four equations-,,, and These, when used in conjunction with the constraints and -, ields, after elimination of the s and and, the rather nast set of two highl non-linear algebraic equations- [ [ and The can be solved graphicall and one finds ,.579.., , and The results also show that the minimum distance between the curves will be δ

5 Note from the above figure that this shortest route between the constraint curves occurs where their slopes are equal, a fact making possible a rapid evaluation of the above equations since the figure allows one to make good initial estimates for the epected values. As a final eample of a Lagrange Multiplier application consider the problem of finding the particular triangle of sides a, b, and c whose area is maimum when its perimeter Labc is fied. Our starting point here is Heron s famous formula for the area of a triangle- A s s a s b s c where s a b c / L / as the half perimeter A little manipulation allows one to recast this result in the form- [ a b c ][ c a b ] A 6 Net appling the Lagrange Multiplier method we have- [[ z ][ z ] [ z] ] or the equivalent three algebraic epressions- [ z ] [ z ], [ z ] [ z ], z[ z ] [ z ]z Eliminating one has U-V-UV--zUVz with Uz -- and V -z. Combining the first two we find -V so that. Combining the second and the last one has zv-uu orzz- impling that z. Thus one ma conclude that the triangle of largest area subjected to the constraint of a fied perimeter is an equilateral triangle. Using the Heron area formula, one has- a A [3a ][ a ] for the largest possible area triangle for fied perimeter L3a.

6 June 9

Lesson 10T ~ Three-Dimensional Figures

Lesson 10T ~ Three-Dimensional Figures Name Period Date Use the table of names at the right to name each solid. 1. 2. Names of Solids 3. 4. 4 cm 4 cm Cone Cylinder Hexagonal prism Pentagonal pyramid Rectangular