Optimization. x = 22 corresponds to local maximum by second derivative test

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Darcy Jenkins
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1 Optimiztion Lectue 17 discussed the exteme vlues of functions. This lectue will pply the lesson fom Lectue 17 to wod poblems. In this section, it is impotnt to emembe we e in Clculus I nd e deling one-vible functions. Clculus III dels with multi-vible functions, but we e inteested fo now in one-vible functions. Accodingly, if we need to find the mximum volume of cylinde given by V = π h, then we need to be ble to substitute fo in tems of h o fo h in tems of so tht the volume fomul becomes function in one-vible. Let's stt off esy nd find two nonnegtive numbes tht dd to 44 such tht the poduct of the two ddends is s lge s possible. Fist, nme the vibles nd ny estictions. We e looking fo two numbes, so we will nme them x nd y nd note tht 0 x 44 nd 0 y 44. econd, wite the function to be optimized in this cse P = xy. Thid, find eltion between the vibles in this cse x+ y = 44 so y = 44 x. Fouth, educe the function to be optimized to one vible using the eltion fom step thee: P x( 44 x) =. Fifth, we find the exteme vlue of inteest using the function fom step fou using eithe the fist deivtive o the second deivtive test. P = 44x x P' = 44 x 44 x 44 = x = x P" = < 0 x = coesponds to locl mximum by second deivtive test Finlly, we use the infomtion fom step five to nswe the question: The two nonnegtive numbes tht dd to 44 such tht thei poduct is s lge s possible e nd (emembe tht y = 44 x nd 44 = ). ("Hey!" The ede sys, " nd e not two numbes." Yes, nd is pi of numbes; they just e not distinct. The poblem did not sy tht the two numbes must be distinct.) Let's summize the steps bove then use the steps to solve nothe stted poblem. 1. Nme the vibles nd estictions.. Wite the function to be optimized. 3. Find eltion between the vibles. 4. Reduce the function fom step two to one vible using the eltion fom step thee. 5. Find the exteme vlues using the fist deivtive o second deivte test. 6. Answe the question.

2 Hee's typicl optimiztion poblem: If closed tin cn in the shpe of ight-cicul cylinde 3 of volume 16 π in, find the height nd dius if the lest mount of mteil is used to mnufctue the tin cn. Accoding to step one, we nme the vibles nd estictions. We need to find height nd dius: h nd, both of which must be positive numbes becuse cn cnnot hve negtive o zeo dimensions. The mteil used is mesued in sufce e, which is given by = π h+ π. Thus, we hve the function to be minimized (step two). We cn use the given volume to find eltion between the vibles (step thee): V = π h 16π = π 16 h = h ubstituting 16 fo h into the sufce e fomul ccomplishes step fou (educing the function to be optimized to one vible). Now, we pefom the second deivtive test: 16 = π + π 1 = 3π + π ' = 3π + π ( ) ( ) ( ) π ( )( + + 4) = + + = + = = = π + 3 '' 64 4 π 6 ''( ) = + 3 '' = 1π > 0 = coesponds to locl minimum by the second deivtive test ince h = 16, then when =, h = 4. Recll tht the units given wee in inches nd stte the nswe: the lest mteil will be used if the cn is 4 inches high nd inches in dimete.

3 Pctice Poblems 1st ed. poblem set: ection 4.6 #3, #5, #13 nd ed. poblem set: ection 4.6 #3 11 odd, #19 3d ed. poblem set: ection 4.6 #3 13 odd, #19 Possible Exm Poblem #1 A box must be mnufctued to hve volume of 88 cubic inches whee the bse is ectngle thee times its width. Wht dimensions of the box equie the lest mteil? Answe: 1 inches 4 inches 6 inches # Detemine the dimensions of the lgest field tht cn enclosed using 500 feet of fencing mteil with one side the field bodeed by ive so tht fencing is not needed on tht side. Answe: 50 feet 15 feet

4 Exmple Execise A ight-cicul cylinde is to be inscibed in sphee of constnt dius. Wht is the tio of the ltitude to the bse dius of the cylinde hving the lgest ltel sufce e ( = π h)? The function hs two vibles. We need to elte the two vibles. Dw digm. Note the ight tingle with the bse dius s one leg, one-hlf the height of the cylinde s the second leg, nd the sphee s dius s the hypotenuse. Wite equtions elting the bse dius nd the height to the ngle between the hypotenuse nd the veticl leg. 1 h sin = = sin nd cos.5h h= cos Rewite the function so tht it depends on one vible. Remembe tht the sphee s dius is constnt. = π sin cos ( )( ) ( ) = π sincos = π sin Tke the deivtive nd find the citicl numbes. ' = cos cos 0 cos = π 4 = Evluting the second deivtive t this citicl numbe yields negtive vlue s below. " = 8π sin ( π ) " 4 = 8π Hence, = π 4 mximizes the function. ubstituting into the equtions bove, we see = nd h=. Theefoe, the tio h=.

5 Appliction Execise uppose tht mnufctue equies tht the luminum contines fo its poduct hve cpcity of 54 cubic inches nd tke the shpe of ight cicul cylinde. Appoximte the dius nd height of the contine tht equies the lest mount of luminum.

Answers to test yourself questions

Answers to test yourself questions Answes to test youself questions opic Descibing fields Gm Gm Gm Gm he net field t is: g ( d / ) ( 4d / ) d d Gm Gm Gm Gm Gm Gm b he net potentil t is: V d / 4d / d 4d d d V e 4 7 9 49 J kg 7 7 Gm d b E