Applications Functions With a Single Critical Number

Transcription

1 Applications 31 Functions With a Single Critical Number Using the CIT we have been able to find the absolute etrema of the continuous functions on a closed interval Using the First Derivative Test, we have found relative etrema for functions on their entire domains, often open intervals However, in man of the applications that we are about to consider, we will be required to find the absolute etreme values of a function on an open interval We can do this easil under the following circumstances THEOREM 311 (SCPT: The Single Critical Point Theorem) Assume that f is differentiable on an interval I (which ma be open) and that b is the onl critical number of f on I 1 If f has a local ma at b, then f actuall has an absolute ma at b If f has a local min at b, then f actuall has an absolute min at b Proof We will do case (1) [Do case () as etra credit] We are given that f has a local ma at = b We want to show that if a is an other point in I, then f (b) > f (a) The first derivative test tells us that f changes sign from positive to negative at b inc l ma dec f This means that f () > 0 b when < b and f () < 0 when > b So suppose that a < b B the MVT there is some point c with a < c < b so that or But f (c) = f (b) f (a) b a f (b) f (a) = f (c)(b a) = (positive)(positive) > 0 f (b) f (a) > 0 f (b) > f (a) which is what we were tring to show Instead, if b < a, then use the MVT again to get f (b) f (a) = f (c)(b a) = (negative)(negative) > 0 and again f (b) > f (a)

2 math 130, das 3 33 applications: ma-min problems EXAMPLE 31 Consider = f () = e which is defined for all or (, ) It has a single critical number: f () = e = 0 at = 0 inc l ma f 0 dec B the SCPT, since f () = e has onl one critical number and it is a relative ma, then it has an absolute ma at the point = 0 3 Applied Optimization (Ma-Min) Problems We now consider a wide variet of optimization problems of the sort we discussed briefl about a week ago Eample A: A soup can is to hold 18 oz What dimensions for the radius and height of the can minimize the cost (ie, the materials) for the can Here, optimize means minimize Eample B: What tuition optimizes (maimizes) the revenue for HWS? Overview Such problems can be readil solved using the following analsis Don t think of this as a recipe to memorize, rather tr to understand the goal of each problem and then make use of the theor we have developed to solve the problem Remember our solution is an essa or argument that in which ou tr to convince our reader (using evidence) that ou have found a solution 1 Identif the problem: Find the absolute ma or min of a quantit Q [that is usuall epressed as a function of two variables, which we call and ] Subject to a constraint: Usuall a constant k = a simple function of and 3 Use the constraint to eliminate a variable, eg, solve for 4 Write Q as a function of one variable, Q(), and state its domain, which is often determined b the constraint equation I cannot overemphasize determining the domain of Q() in this process 5 Find the absolute ma or min of Q on its domain The method ou use will be determined, in part, b the domain If the domain is a closed interval, ou can use the CIT If it is not closed, then ou must hope that there is onl one critical number so that ou can use the SCPT (The SCPT can also be used on a closed interval if there is a single critical number) Justif our answer in a sentence Notice that onl in step (5) do we actuall do an calculus 33 Lots of Eamples EXAMPLE 331 A real estate developer wishes to enclose a rectangular plot b a fence and then divide it into two plots b another fence parallel to one of the sides What are the dimensions of the largest area that can be enclosed b using a total of 1800 meters of fencing?

3 33 LOTS OF EXAMPLES 3 Solution Let s use the guidelines to attack the problem 1 Maimize A = Subject to P = 1800 = Eliminate = Notice that since the smallest can be is 0 if none of the fence is used for the -sides; the largest can be is 600 feet if all of the fence is used for the three -sides (since = 1800) 4 Substituting for, write A in terms of a single variable with its domain: A() = (900 3 ) = on [0, 600] 5 Solve the problem Use CIT A = = 0 at = 300 Check the critical points and endpoints to determine the absolute ma A(300) = 300(450) = 135, 000 square feet A(0) = 0 and A(600) = 0 B the CIT, A is maimized when = 600 and = 450 A = 135, 00 square feet EXAMPLE 33 Find the dimensions of the bo with a square base that has a volume of 51 in 3 and has the smallest surface area (4 sides, top and bottom) Solution Using the guidelines h 1 Minimize surface S = 4h + Subject to V = 51 = h 3 Eliminate h We get h = 51 Notice that must be greater than 0 but can be as large as we like (we can still solve for h, though it gets smaller as gets larger) So the domain is (0, ) 4 Rewrite S S = = on (0, ) 5 Since the interval is not closed, we must use the SCPT, if it has onl one critical number S = = 0 4 = = 51 = 8 Yes, there is onl one critical point, so now classif it f inc l min dec Since there is onl one critical point and it is a local min, b the SCPT it is also an absolute min So the surface area is minimized when = 8 and h = 51 8 = 8 The minimum surface area is S = 4(8)(8) + (8 ) = 384 square inches

4 math 130, das 3 33 applications: ma-min problems 4 EXAMPLE 333 A cable television compan has its master antenna located at point A on the bank of a straight river 1 kilometer wise (Figure 1 above) It is going to run a cable from A to a point P on the opposite bank of the river and then straight along the bank to a town T situated 3 kilometers downstream from A It costs $15 per meter to run the cable under the water and $9 per meter to run the cable along the bank What should be the distance from P to T in order to minimize the total cost of the cable? A 1 km Solution Q P 3 T Using the guidelines 1 Minimize the cost C = (3 ) Subject to = Eliminate Here, = + 1, where from the diagram Rewrite C C() = (3 ) on [0, 3] 5 Use CIT (or possibl SCPT) Find the critical numbers C () = = 0 15 = = 81( + 1) This means 144 = 81 1 = ±9 = ±3/4 Since 3/4 is not in the interval, so there is onl a single critical point at = 3/4 Use the CIT C(3/4) = 15 5/16 + 9(9/4) = 15(5/4) + 9(9/4) = 39 C(0) = = 4 C(3) = So b the CIT, the absolute min occurs when = 3/4 and the cost is 39 EXAMPLE 334 A child s sandbo is to be made b cutting equal squares from the corners of a square sheet of galvanized iron and turning up the sides If each side of the sheet of galvanized iron is meters long, what size squares should be cut from the corners to maimize the volume of the sandbo? Solution This time 1 Maimize the volume V = Subject to = + where Eliminate =

5 33 LOTS OF EXAMPLES 5 4 Rewrite V V() = ( ) = on [0, 1] 5 Use CIT (or possibl SCPT) Find the critical numbers V () = = 4( ) = 4(1 )(1 3) = 0 = 1, 1/3 Use the CIT V(1/3) = 1 3 ( 4 3 ) = 16 7 V(0) = 0 = V(1) So b the CIT, the absolute ma occurs when = 1/3 and the volume is 16 7 cubic meters EXAMPLE 335 Advertising fliers are to be made from rectangular sheets of paper that contain 400 square centimeters of printed message If the margins at the top and bottom are each 45 centimeters and the margins at the sides are each centimeters, what should be the dimensions of the fliers if the total page area is to be a minimum? Solution This time 1 Minimize the total area A = ( + 4)( + 9) Subject to constraint = Eliminate = 400, domain (0, ) 4 Rewrite A A() = ( + 4)( Use SCPT Find the critical numbers + 9) = on (0, ) A () = = 0 = 1600 = ± Onl 40/3 is in the domain, so we can use SCPT Classif the critical point using the First Derivative Test f dec l min inc /3 Since there is onl one critical point and it is a local min, b the SCPT it is also an absolute min So the area is minimized when = 40/3 The flier dimensions should be + 4 = = in and + 9 = 40/3 + 9 = 39 in EXAMPLE 336 In New Geneva, two straight roads and a highwa mark off a triangular piece of land (Figure above) The side of the triangle ling along the highwa is 100 meters long, and the perpendicular distance from the intersection of the two roads to the highwa is 80 meters Within this triangle, the cit intends to zone a rectangular plot of land,

6 math 130, das 3 33 applications: ma-min problems 6 fronting along the highwa, for commercial use The park department will plant shrubs in the remaining portion of the triangle What is the maimum possible area of the rectangular plot of commerciall zoned land? Solution 80 h h w 100 m 80 m 1 Maimize A = hw Subject to the constraint (similar triangles): w 100 = 80 h 80 = Eliminate w w = h on [0, 80] 4 Rewrite A = ( h)h = 100h 5 4 h on [0, 80] 5 Use either CIT or SCPT if possible A = h = 0 h = 40 Use the CIT: Check the endpoints: A(0) = 0, A(80) = 0 and at the critical point A(40) = ( (40))(40) = 000 square ft This the absolute ma EXAMPLE 337 (A Ke Ma-Min Problem) Optical fiber from a computer lab to a phsics lab is being laid The phsics lab lies across a 15 m wide road and is 5 m down the other side from the computer lab (See figure) It costs $13(00) per meter to put cable under the road and $1(00) per meter to bur it underground Where should the cable be brought across the road? Phs Lab 15 m Solution Q P 5 CS Lab 1 Minimize the cost C = (5 ) Subject to = + (15) 3 Eliminate = + 65 where Rewrite C C() = (5 ) on [0, 5] 5 Use CIT (or possibl SCPT) Find the critical numbers C () = This means = 0 13 = = 144( + 5) 5 = (144)(5) 5 = ±1(15) = ±36 Since NEITHER point is in the interval Use the CIT The min occurs at an endpoint Using C() on [0, 5], we get: C(0) = (5 0) = 13(15) + 1(5) = 495 C(5) = (5 5) = = So b the CIT, the absolute min occurs when = 0 and the cost is $37,901

7 33 LOTS OF EXAMPLES 7 EXAMPLE 338 Find the dimensions of the rectangle of largest perimeter that can be inscribed in a circle of radius r r θ Solution 1 Maimize the perimeter P = + Subject to = r sin θ and = r cos θ Remember r is constant 3 Eliminate both and! 4 Use Use P = (4r cos θ)(4r sin θ) = 16r cos θ sin θ where θ is in the interval [0, /] 5 Use CIT (or possibl SCPT) P = 16r ( sin θ + cos θ) = 0 cos θ = sin θ cos θ = ± sin θ The onl angle in [0, /] where this is true is when θ = /4 Use CIT At the critical point: P(/4) = 16r cos(/4) sin(/4) = 16r ( )( ) = 8r At the endpoints: P(0) = 16r cos(0) sin(0) = 16r (1)(0) = 0 and P(0) = 16r cos(/4) sin(/4) = 16r (0)(1) = 0 So the absolute ma is 8r at θ = /4 EXAMPLE 339 US Postal regulations state: Priorit Mail is used for documents, gifts, and merchandise The maimum size is 108 inches or less in combined length and distance around the thickest part (girth)" Find the dimensions of the clinder of largest volume that can be sent Priorit Mail Solution 1 Maimize V = r h Subject to Girth + length = 108 = r + h 3 Eliminate h h = 108 r on [0, 54 ] 4 Rewrite V = r (108 r) = 108r r 3 on [0, 54 ] 5 Use CIT (or perhaps SCPT) Find the critical numbers V = 16r 6 r = 6r(36 r) = 0 r = 0, Use CIT Check V at the endpoints and critical numbers At the endpoints: V(0) = 0 and V( 54 ) = 0 At the critical number: V( 36 ma occurs at r = ) = ( ) (108 ( 36 )) > 0 So the absolute ) = = 36 and h = 108 ( 36 EXAMPLE 3310 Find the dimensions of the clinder of largest surface area that can be sent Priorit Mail Solution bottom) This time maimize A = rh + r (the tube plus the top and

8 math 130, das 3 33 applications: ma-min problems 8 As before, h = 108 r on [0, 54 ] So A = r(108 r) + r = 16r 4 r on [0, 54 ] So A = 16 8 r = 8(7 r) = 0 r = 7 Let s use SCPT Justif: Use the first derivative test: A there is a relative ma at r = 7 B the SCPT there is an absolute ma there h = 108 ( 7 ) = = 54 EXAMPLE 3311 Soda cans hold 113 cm 3 (355 ml) The tops and bottoms are made of aluminum that is double thick Find radius of the can that minimize the material used to make (sides, top, and bottom of) the can How does this compare to the radius of an actual soda can? Solution This time minimize A = rh + (r ) (the tube plus double top and bottom) The constraint is V = 113 = r h Eliminate h = 113 on (0, ) r So A = r( 113 ) + 4r = 6 r r + 4r on (0, ) We epect to use SCPT A = 6 r + 8r = 0 8r = 6 r r 3 = r = 3 4 Use SCPT First classif the critical number Use the first derivative test: 7 A ( ) 1/3 is a relative min at r = B the SCPT there is an absolute min there Measuring a real can we see that the radius is approimatel 315 cm Comparing EXAMPLE 331 A Norman window is a window that consists of a rectangle surmounted 1 b a semicircle These windows were commonl constructed in English architecture for about 100 ears after the Norman conquest of Britain, circa ) If the perimeter is 4 ft, find the radius that will maimize the area of the window 1 This ma be one of the few remaining uses in English of the word surmount" in reference to phsical objects r Solution Maimize: Area A = r + 1 r (rectangle plus semicircle) Subject to the constraint: 4 = + r + r (circumference) Eliminate: = 1 r 1 r = 1 ( )r with domain: 0 r 1+(/) Rewrite: A = r(1 [1 + 1 ]r) + 1 r 1 on [0, 1+(/) ] So A = 4r ( + )r + 1 r = 4r ( + 1 )r

9 33 LOTS OF EXAMPLES 9 Differentiate: A = 4 (4 + )r = 0 r = 4 4+ Justif: Use the first derivative test: A R Ma at r = 4+ 4 Since there is onl one critical number, there is an absolute ma at r = 4+ 4 b SCPT EXAMPLE 3313 A clindrical waste basket (no top) has a volume of 1000 cubic inches What radius will minimize the material (surface area) of the basket 4 4+ Solution 1 Maimize S = r + rh (bottom circle plus side) Subject to V = 1000 = r h on (0, ) 3 Eliminate h h = 1000 r 4 Rewrite S = r + (r) 1000 r = r r 5 Use SCPT Find the critical numbers S = r 000 r on (0, ) = 0 r = 000 r r 3 = 1000, r = 10 Use SCPT First classif the critical number Use the first derivative test: S is a relative min at r = 10 B the SCPT there is an absolute min there 10