Calculus 1 Optimization Problems


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1 9 Calculus Optimization Poblems ) A Noman window has the outline of a semicicle on top of a ectangle as shown in the figue Suppose thee is 8 + π feet of wood tim available fo all 4 sides of the ectangle and the semicicle Find the dimensions of the ectangle (and hence the semicicle) that will maimize the aea of the window ) You ae building a clindical bael in which to put D Bent so ou can float him ove Niagaa Falls I can fit in a bael with volume equal cubic mete The mateial fo the lateal suface costs $8 pe squae mete The mateial fo the cicula ends costs $9 pe squae mete What ae the eact adius and height of the bael so that cost is minimized? ) A ectangula sheet of pape with peimete 6 cm is to be olled into a clinde What ae the dimensions of the sheet that give the geatest volume? 4) A ight tiangle whose hpotenuse is m long is evolved about one of its legs to geneate a ight cicula cone Find the adius, height, and volume of the cone of geatest volume Note: V = π h h 5) Detemine the clinde with the lagest volume that can be inscibed in a cone of height 8 cm and base adius 4 cm
2 9 Calculus Optimization Poblems 6) A staight piece of wie 8 feet long is bent into the shape of an L What is the shotest possible distance between the ends? 7) Find the dimensions of the ectangle of lagest aea that has its base on the ais and its othe two vetices above the ais and ling on the paabola = 9 8) A closed clindical containe is to have a volume of 00 π in The mateial fo the top and bottom of the containe will cost $ pe in, and the mateial fo the sides will cost $6 pe in Find the dimensions of the containe of least cost a)daw a pictue, label vaiables and wite down a constained optimization poblem that models this poblem (5 Pts) b) Using calculus, solve the poblem in pat (a) to find the dimensions 9) A closed ectangula containe with a squae base is to have a volume of 00 in The mateial fo the top and bottom of the containe will cost $ pe in, and the mateial fo the sides will cost $6 pe in Find the dimensions of the containe of least cost 0) You deam of becoming a hamste beede has finall come tue You ae constucting a set of ectangula pens in which to beed ou fu fiends The oveall aea ou ae woking with is 60 squae feet, and ou want to divide the aea up into si pens of equal size as shown below The cost of the outside fencing is $0 a foot The inside fencing costs $5 a foot You wish to minimize the cost of the fencing a) Labeling vaiables, wite down a constained optimization poblem that descibes this poblem b) Using an method leaned in this couse, find the eact dimensions of each pen that will minimize the cost of the beeding gound What is the total cost?
3 9 Calculus Optimization Poblems Solutions: ) We will assume both and ae positive, else we do not have the equied window Let P be the wood tim, then the total amount is the peimete of the ectangle the cicumfeence of a cicle of adius, o π Hence the constaint is The objective function is the aea P = π = 8 + π A = + π 4 + plus half Solving the constaint fo gives 8 + π (4 +π ) = and so A = ( 8 + π (4 + π ) ) + π o A = + π + π + π 8 + π And so we wish to maimize A ove the inteval 0, 4 + π d A = ( 8 + π ) (4 + π ) + π d ( 8 ) (4 ) d A Which is 0 when = Since = ( 8 π ) < 0, we have indeed have a maimum d Since = implies =, the dimensions of the ectangle ae b feet Student ma choose the altenative wa to solve the poblem Assume is a function of and using implicit diffeentiation: π = 8 + π and A = + π
4 9 Calculus Optimization Poblems o so d d d A d ( π ) = 0 and = ( + π ) d d d π = 0 d and d A d = + +π d d d d 4 + π = d and + + π = 0 d 4 + π + π = 0 Using this in the oiginal constaint eq gives o = π = (8 + π ) = 8 + π So =, and =, and the dimension of the ectangle is b feet +=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+= ) The volume is given b V = π h The cicula ends have total aea π The lateal suface aea is π h Cost is C = ($8)(π h) + ($9)(π ) h So the poblem at hand is minimize C = 6π h + 8π subject to the constaint π h = π h = means h = Using this hee gives π and so 6 C = + 8π C = 6 6π +, which gives citical points as = 0, and = π
5 9 Calculus Optimization Poblems The phsicall easonable solution is = m which gives π h = m π Note: If ou wish to solve the poblem using implicit diffeentiation The steps follow h The volume is given b V = π h The cicula ends have total aea π The lateal suface aea is π h Cost is C = ($8)(π h) + ($9)(π ) So the poblem at hand is minimize C = 6π h + 8π d d subject to the constaint π h = Assuming h = f (): d d d ( π h) = () and C = (6π h + 8π ) d d d dh dc d h π h + π = 0 and = 0 = 6π h + 6π + 6π d d d which gives d h h = d h 6 π h + 6π + 6π = 0 o = h Using = h in π h = gives = h = m π +=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=
6 9 Calculus Optimization Poblems ) Let and be the dimensions of the sheet of pape Since + = 6, + = 8 is the constaint The adius is given b π =, so =, and the volume is V = π = Using π 4π (8 ) dv 6 =8, V = is the function to be optimized =, so citical numbes 4π dt 4π ae = 0, Maimum volume occus when =, (Wh?) so dimensions ae 6 cm b cm 6 and the volume is cm π +=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+= 4) The volume of the cone is: V = π h h The constaint equation is: + h = ( ) o + h = ) Solving fo gives = h, so V ( h ) h = π o V = π (h h ) dv Taking the hdeivative of V gives = π ( h ) = π ( h ) dh Stationa points ae h = ±, and the phsicall easonable one is h = If h =, then = h = =, so =, and the volume is V = π ) Solving fo h is FAR moe difficult + h = means h =, and so V dv So = π d Rewiting this as dv = π d + π + π π 6 = ( ) = = π So stationa points ae = 0, ±, ± The onl phsicall easonable one is =, so h = =, and V = π
7 9 Calculus Optimization Poblems +=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+= 5) Detemine the clinde with the lagest volume that can be inscibed in a cone of height 8 cm and base adius 4 cm (5 points) (0, 8) (, ) (4, 0) The volume of the clinde is V = π The equiement that the clinde is inscibed in the cone leads to the pictue on the ight Since the edge of the cone is a staight line, we can use the 8 0 two points (0, 8) and (4, 0) to detemine the elation between and Slope is = The 0 4 line is = + 8 So the volume becomes V = π (8 ) = π (8 ) and [0,4] V = π (6 6 ) = π (8 ) 8 8 And so citical numbes ae = 0, The maimum occus when the adius is = which 8 5π means the height is =, and the volume is V = cubic cm 7 6)A staight piece of wie 8 feet long is bent into the shape of an L What is the shotest possible distance between the ends? (0 Points) d
8 9 Calculus Optimization Poblems Let d be the distance between the ends of the L then and so = 8, and then d d + = The constaint is + = 8 = f ( ) = + (8 ) which, when simplified, gives, f ( ) = Since d and d have the same citical points we wok with f () To detemine citical numbes we compute f ( ) = 4 6, and solve 4 6 = 0, and so the citical numbe is = 4, and so = 8 = 4 also The distance is then d = + = = 4 We know this is a minimum since f ( ) = 4 > 0 7) Find the dimensions of the ectangle of lagest aea that has its base on the ais and its othe two vetices above the ais and ling on the paabola = 9 = 9  ^ The ectangle aea is A = The equiement that the ectangle lies on the gaph of = 9 means A = (9 ) = 8 The vaiable is esticted between [0,] at which both points ield a minimum value of no ectangle o 0 aea Since When A ( ) = 8 6 = 6( ), A ( ) = 6( ) = 0 we get
9 9 Calculus Optimization Poblems citical points at = ± Fo ou geomet we choose the positive oot =, and so, = 6, so the dimensions ae units b 6 units, and the aea is = squae units 8) h Volume: V = π h Cost: $(π ) + $6(π h) So Poblem is minimize costc = 4π + π h subject to the constaint V = π h = 00π and so h = b) Solving this last equation fo h gives: h =, which when substituted into the cost equation 600π ields C = 4π + The geomet gives ( 0, ) To minimize the cost we detemine 600π citical numbes fom C = 8π = 0 hence = 450 so the citical numbe is / / 00 (450) 700 = (450) This gives h = = in Since C = 8π + π = 0π > 0, the / (450) / dimensions ield the minimum cost The clinde should have a adius = (450) in, and a 00 height of h = in ode to minimize the cost / (450) 9) A closed ectangula containe with a squae base is to have a volume of 00 in The mateial fo the top and bottom of the containe will cost $ pe in, and the mateial fo the sides will cost $6 pe in Find the dimensions of the containe of least cost (0 Points) Volume: V = h Cost: $( ) + $6(4 h) So Poblem is minimize C = h subject to the constaint h = 00 Solving this last equation fo h gives: h 00 h =, which when substituted into the cost equation 700 ields C = 4 + The geomet gives ( 0, ) 700 Since C = 8 = 0 gives = 900 whose solution is = 900 / 9 65 This gives / / / h = = in Note h = 900 = = 00 in so the volume is / C = 4 +
10 9 Calculus Optimization Poblems 4,400 coect And since C = 8 + = > 0, the dimensions ield the minimum cost The bo / 900 / should have a squae base of side length 900 in, and a height of in FYI 600 C () ) The cost of the outside fencing is $0 a foot The inside fencing costs $5 a foot You wish to minimize the cost of the fencing a) Let be the width of each individual pen, and be the length as shown above Since the total aea is 60 sq ft, each individual pen will have an aea of 0 sq ft The constaint is = 0 The objective function is the cost Eamining the fencing above, thee is 5 feet of inteio fencing, and + feet of eteio fencing So the total cost is C = $ 5 (5) + $0 ( + ), o C = The constained optimization poblem is: Minimize =0 C = subject to the constaint b) Solving = 0 of gives = Substituting this into C gives C = + 0, as the function to minimize ove ( 0, ) 450 C = + 0, and so citical points ae = 0, and = = The cost is C = = 0 5 $ = 5, so
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