The Krewe of Caesar Problem. David Gurney. Southeastern Louisiana University. SLU 10541, 500 Western Avenue. Hammond, LA

Transcription

1 Te Krewe of Caesar Problem David Gurney Souteastern Louisiana University SLU 10541, 500 Western Avenue Hammond, LA 7040 June 19, 00

2 Krewe of Caesar 1 ABSTRACT Tis paper provides an alternative to te usual max-min problems found in calculus. In addition to combining many matematical ideas to reac a solution, te problem presented as te added advantage of dealing wit a topic of regional interest to tose in Louisiana and Mississippi. Te Krewe of Caesar is a Mardi Gras Parade in te New Orleans suburb of Metairie, LA. During te course of te parade, te floats and bands turn off of Veterans Memorial Blvd. and onto Bonnabel Blvd. Te parade travels along te westbound lanes for a wile and ten, aving no place else to go, returns to Veterans along te eastbound lanes. Parade goers wo are on median (or neutral ground in te native parlance) of Bonnabel Blvd. can run back and fort, catcing beads and oter items (collectively know as trows) from floats traveling bot down te eastbound and westbound lanes. Tis is te source of te problem. Jon and Melissa are bot parade goers wo ave staked out a portion of te neutral ground on Bonnabel Blvd. Melissa stands in one spot guarding Jon s cace of beads wile e

3 Krewe of Caesar runs back and fort across te median to te floats traveling by on eiter side. Jon is next to Melissa as two floats, Trojan and Viking, approac in opposite directions on Bonnabel Blvd. Te figure above depicts tis situation. Jon wants to run to one side and catc trows from Trojan as it passes, ten run to te oter side and catc trows from Viking as it passes, and ten return wit is ard-won tropies to Melissa. Te problem is ow to do tis so tat te distance Jon travels is a minimum. If M is Melissa s location, T is te Trojan catc spot, and V is te Viking catc spot, ten te object ten is te minimize te sum (1) dist(m,t) + dist(t,v) + dist(v,m). For te solution presented in tis paper, some simplifying assumptions are now made. Assume tat Jon s speed is constant and te two float speeds are constant and equal. Also assume tat Jon can gauge te speed of a float instantaneously just by watcing it move. Wit tese assumptions, one sould note tat te location of T determines te total distance Jon will ave to run. Tat is, after catcing trows from Trojan, Jon must travel in a given direction to meet te leading edge of Viking. Having met te leading edge, Jon will stop to catc trows as Viking passes and ten e must return directly to Melissa. Coosing te direction to travel from T to V amounts to solving te following. Suppose tat P and Q are projectiles wic are moving wit constant speeds v and w, respectively. Initially, P and Q are positioned as sown in te figure below so tat te point O is te vertex of a rigt angle containing P and Q. Te distance from P to O is a, te distance from O to Q is b, and Q is traveling parallel to te segment PO. Te angle between te segment PO and P s direction of travel must be cosen so tat objects P and Q will meet at point R at some time t after te start.

4 Krewe of Caesar 3 Given tese conditions, te following equations must be satisfied. () tvcosθ + tw= a tvsinθ = b Note tat sin must be positive wile cos could be negative. Tus tere is just one angle between 0 and 180 wic will satisfy tese conditions. Solving for sin and converting to cosine gives (3) tv b cos θ =±. tv Substituting tis into te first equation of () gives (4) ± tv b = a tw. Squaring bot sides results in a quadratic equation wit te solution (5) ( ) aw± v a + b w b t = v w. Since t is positive, te plus solution is cosen. Multiplying t by v ten gives (6) ( ) avw+ v v a + b w b Dist( P, Q) =. v w Tese formulas provide te basis for te solution to te original problem. To keep te solution presented ere fairly simple, specific values will be assigned to te constants before proceeding. Let Jon s speed be 10 feet per second. Suppose bot Trojan and

5 Krewe of Caesar 4 Viking travel at 5 feet per second and are eac 0 feet long. Finally, let te neutral ground be 40 feet wide. By placing M at te center of te neutral ground, a coordinate system can be set up wit te origin at M and te orizontal axis parallel to Bonnabel Blvd. In tis situation, te coordinates of T and V would be (, 0) and (k, 0), respectively. Te values of and k are unknown at te moment, but tey can be found using te formulas (5) and (6). Te figure below diagrams Jon s route. Te formula for te distance traveled in te first leg of Jon s trip results from a simple application of te Pytagorean Teorem. (7) dist M T (, ) = Te distance traveled in te second leg requires knowing te position of Viking wen Jon leaves te Trojan catc spot. Assuming tat Jon waits at te Trojan catc spot wile te full lengt of Trojan passes, te leading edge of Viking will ave moved to te point

6 Krewe of Caesar 5 ( 0, 0). Tus in formulas (5) and (6), a = ( 0) ( ) = 0, b = 40, v = 10 and w = 5. Tis gives 1 (8) Time T V = ( + + ) and multiplying by te velocity v, (, ) , 15 (9) Dist T V = ( + + ) (, ) Te catc spot for Viking will be at ( 0 5 Time(T,V), 0). Given tese coordinates, te Pytagorean Teorem is used again to find te lengt of te final leg. 1 (10) Dist( V, M ) = ( ) 9 Te total distance Jon travels, te sum of te tree distances (7), (9) and (10), is a function of te just one variable,. Usually one takes te derivative at tis point and sets it equal to zero. Unfortunately, in tis case, taking te derivative leads to an expression wit a few different radical expressions, and small ope of finding its roots. On te oter and, graping te function is fairly easy. Te grap below, constructed using Matematica, sows tat te minimum distance of about 97 feet occurs wen is about 0 feet. Te problem is ow to verify tat tis computer-generated image is accurate. Total Distance /

7 Krewe of Caesar 6 One option is to analyze te distance components, (7), (9) and (10), individually. At first, even tis does not look too promising, but some useful observations can be made. For one ting, Dist(M, T) decreases for < 0 and increases for > 0, as sown in te grap below and confirmed by te derivative. Dist H M,TL (11) d Dist( M, T ) = d For anoter, in te last two components, (9) and (10), te expression under te radical, , is always positive. Tis fact can be used to sow tat te derivative (1) d 4 0 Dist( T, V ) = 1 d decreases for < 0 and increases for > 0. Tis result is supported by te grap below. Dist H T,VL

8 Krewe of Caesar 7 Finally, te grap below sows tat Dist(V, M) is always decreasing. Verifying tis requires a closer look at te formula (10). Only te expression being squared, (13) , needs to be considered. Tis expression is always negative and always increasing. Terefore its square is always positive and always decreasing. Hence Dist(V, M) is always decreasing. Dist H V,ML Tus, putting tese conclusions togeter, te only place for te minimum total distance to occur is somewere near = 0. Tere are many possible variations on tis problem. To name just a few, one could (a) give te floats different speeds, (b) let Jon accelerate as e runs, or (c) let te effective lengts of te floats cange during te course of te parade. Variation (c) would bring into consideration te fact tat float riders become tired as te parade moves along. As a result, te distance te beads are trown decreases in some fasion, and ence te spread of people along te curb of Bonnabel wo catc trows from te float at a single spot i.e. te effective float lengt - decreases as well.

9 Krewe of Caesar 8 Anoter consideration in tis problem is te existence of solutions. Will tere always be a solution if Jon runs at least as fast as te floats? Under wat conditions could Jon run at te same or a slower speed tan te floats and still reac tem bot? Ten tere are optimization problems. Suppose te trows from one float are more valuable tan te trows from te oter float. Wat route sould Jon take ten to maximize te value of is take? Tese are just a few lines of investigation, and tere are bound to be more. Wit so many questions to answer, te Krewe of Caesar problem could provide te basis for a senior project, or maybe even a master s tesis. On te oter and, it migt provide some recreation to tose wo are burnt out on oter problems. In any case, ave fun.