Math 259 Winter Solutions to Homework #9

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1 Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier equtions "f = #"g: = λ y = λy z = λz. We will rek the solution of these three equtions into three cses. Cse :, y nd z re ll nonzero. In this cse, we cn simplify the system of three equtions to get: = y = z = λ/. Sustituting this into g(, y, z) = nd solving gives tht = y = z = /. This mens tht there re totl of eight points tht hve, y, nd z ll non-zero, ech of which looks like: At ech such point, f(, y, z) = /. (,y,z) = ( ±,±,± ). Cse : Ectly one vrile (, y or z) is equl to zero. For the ske of demonstrtion, let us suppose tht = nd tht neither y nor z is zero. Then the Lgrnge multiplier equtions give tht: y = z = λ/. Sustituting these vlues into the constrint eqution g(, y, z) = nd solve for y nd z gives collection of four points, ech of which hs the form: (,y,z) = (,±,± ). There re two dditionl fmilies of points like this, one corresponding to y = nd the other corresponding to z =. At ll of the points in ll of the three fmilies identified here, f(, y, z) = ½.

2 Cse : Ectly two vriles re equl to zero. For the ske of demonstrtion, let us suppose tht = nd tht y =, nd tht z is not zero. Sustituting these vlues into the constrint g(, y, z) = give us the solution z = ±. At ech solution point, f(, y, z) =. Solution of prolem: The mimum vlue of f(, y, z) is nd the minimum vlue of f(, y, z) is /.. The Lgrnge multiplier equtions for the function: suject to the constrint re ll of the form: f (,,..., n ) = n g(,,..., n ) = n "= = λ j with j =,, n. If λ =, then none of the Lgrnge multiplier equtions cn e solved, so we will ssume tht λ is not equl to zero. This gives tht: = = = n. Sustituting the fct tht ll of the vriles must e equl t the solution points into the constrint g(,, n ) = gives tht for ech j =, n: j = ± n. Then t ech solution point, the vlue of the function f(,, n ) will e sum of n copies of the j given ove, with some positive nd some negtive. The mimum vlue of f(,, n ) will e ttined when ll of the terms dded together re positive, nd will produce mimum vlue of f(,, n ) equl to n. The minimum vlue of f(,, n ) will e ttined when ll of the terms dded together re negtive, nd will produce minimum vlue of f(,, n ) equl to n.. In this prolem we re sked to minimize the cost function:

3 C = ml + nk, where L nd K re suject to the constrint L α K - α = Q. This clcultion is one of the most eutiful pplictions of the Lgrnge Multipliers technique in ll of economics. We will write the functions involved in the fmilir Lgrnge Multiplier nottion s: f(l, K) = ml + nk g(l, K) = L α K - α Q. The two grdients tht we need to clculte re s follows: "f =< m,n > "g =< #L #$ K $#,( $# )L # K $# >. The Lgrnge multiplier equtions, "f = #"g, re: m = "#L #$ K $# n = "( #$ )L $ K #$. Multiplying the first eqution y L nd the second eqution y K nd mking the sustitution L α K - α = Q into oth gives: ml = "#Q nk = "( #$ )Q. Solving oth equtions for λq nd equting llows us to solve for L in terms of K to get: L = n" m( #" ) K. We will now use this eqution to sustitute for L in the constrint eqution g(l, K) = nd solve for K. Doing this gives: ( K = Qm" #" ) " nd L = n " " " Qn "# # "# ( ) "#. m "# "# Who knew the disml science of economics could e so pretty!

4 Prolems from Pges (Section.). The region [, ] [, ] is divided up into eight squres ech with width Δ = nd height Δy = s shown in the digrm (elow). y - - Ech of the little squres is the se of French fry. To get the height of ech French fry in this prolem, we hve to evlute the function t the top left-hnd corner of ech squre (indicted y the dots in the digrm given ove). Doing this gives tht the volume is pproimted y: Volume f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy = =.. In this prolem the heights of the French fries re given y the function: f(, y) = + y. The region [, ] [, ] of the y-plne is divided into four smller rectngles with width Δ = nd height Δy = s shown in the digrm (elow). These smller rectngles form the ses of the French fries. The dots on the digrm indicte where the function f(, y) is evluted in Prt () nd Prt () of this prolem.

5 5 y - () In this cse, the function is evluted t the ottom, right-hnd corner of ech rectngle. Volume f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy + f(, ) Δ Δy = ( ) =. () In this cse, the function is evluted t the center of ech rectngle. Volume f(.5, ) Δ Δy + f(.5, ) Δ Δy + f(.5, ) Δ Δy + f(.5, ) Δ Δy = ( ) = 88. (c) The doule integrl tht gives the ect vlue of the volume is: Volume = " " ( + y )ddy = " [ + y ] dy

6 = " ( + y )dy = y + y [ ] = 8/. 6. There re mny, mny wys to set up nd clculte n estimte for the volume of the pool, given the informtion supplied in this prolem. The nswer or estimte tht you otined for the volume of the pool my e little different (nd mye little more ccurte) thn the one given here, ut your numer should e resonly close to the numer listed elow. Let f(, y) give the depth of the pool (in meters) meters to the right nd y meters ove the lower left hnd corner of the pool (which corresponds to the depth in the upper left hnd corner of the tle given on pge 67). We will rek the volume of the pool up into French fries with ses tht hve width Δ = nd height Δy =, s shown in the digrm shown elow. (The dots re the points where the function is evluted to otin the height of ech French fry.) y Then the volume, V, of the pool will e pproimted y: V f(5, 5) Δ Δy + f(5, 5) Δ Δy + f(5, 5) Δ Δy +

7 f(5, 5) Δ Δy + f(5, 5) Δ Δy + f(5, 5) Δ Δy = ( ) = 6 cuic feet.. # # ( + y )dyd = # y + y d = # ( + )d = + " [ ] " [ ] = 6.. The shdow cst in the y-plne y this volume is the rectngle [, ] [, ]. The volume elow the surfce z = + ( ) + y over this rectngle is given y the doule integrl: # # ( + ( ") + y )dyd =. Prolems from Pges (Section.) 8. The doule integrl is given y: y " " + dyd = y # & " % $ ( d = + ' 8 " + d = 8 ln + [ ( )] = 8 ln ( ).. The doule integrl is given y: " " ( + y)dyd = " [ y + y ] d = " ( / + # # )d = 5 / 5 + # # 5 [ ] =

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