Example Optimization Problems selected from Section 4.7

Transcription

1 Example Optimization Problems selecte from Section ) We are aske to fin the points ( X, Y ) on the ellipse 4x 2 + y 2 = 4 that are farthest away from the point ( 1, 0 ) ; as it happens, this point is also on the ellipse, but that is actually of no importance to the solution. We can use the istance formula to calculate the separation of a hypothetical farthest point from the point ( 1, 0 ) from s 2 = ( X 1 ) 2 + (Y 0 ) 2. Since the point ( X, Y ) is on the ellipse, we can write 4X 2 + Y 2 = 4 Y 2 = 4 4 X 2. When we insert this into the istance equation, we obtain s 2 = ( X 1 ) 2 + Y 2 = ( X 1 ) 2 + (4 4X 2 ) = X 2 2X X 2 = 5 2X X 2. It is the istance s we wish to minimize, but since it is a positive number, minimizing s 2 will serve as well. We can thus ifferentiate s 2 with respect to X an set that erivative equal to zero, in orer to locate critical numbers: X (s2 ) = X (5 2X X 2 ) = 2 6X = 0 6X = 2 X = 1. We can now put this critical number into the equation for the ellipse in orer to fin the Y-coorinates of the most istant points: Y 2 = 4 4 X 2 = 4 4 ( 1 ) 2 = = 6 4 = Y = ± 9 = ± 2 16 = ± 4 2 ± (continue)

2 There are two symmetrically-place most istant points at ( 1, 4 2 ) an ( 1, 4 2 ), as we woul expect from the geometry of the ellipse; what is perhaps surprising is that the farthest points are not at the ens of the long axis of the ellipse. 2 We see that the secon erivative is X 2 (s2 ) = 6 < 0, so the critical point for the istance-square (an thus the istance) is inee a maximum. 27) A right circular cyliner is one for which the cross-sections are circles an the axis through the centers of those circles is perpenicular to the base (in fact, what most people mean by a cyliner ). Having this cyliner inscribe within a sphere means that the eges of the top an bottom ens of the cyliner are in contact with the insie of the sphere. We wish to fin the cyliner with the largest volume that can fit insie a sphere of raius r. If we were to chop through the cyliner an the sphere along the symmetry axis of the cyliner, we woul see a cross-section with a rectangle touching a circle at each of the rectangle s corners. We can place one of these corners at the point ( x, y ), which woul put the other corners at ( x, y ), ( x, y ), an ( x, y ). This woul give our inscribe cyliner a base raius of x an a height of 2y, making its volume V = π x 2 (2y ) = 2π x 2 y ; this is the function we wish to maximize. We o know one other thing: since the corners of the rectangle contact the circle of raius r (the raius of the sphere), we have a constraint equation which tells us the relationship between the values of x an y, x 2 + y 2 = r 2. We coul use the equation for the circle to eliminate either x or y ; if we choose to eliminate x, though, we can avoi having to ifferentiate a square root: x 2 = r 2 y 2 V = 2π x 2 y = 2π (r 2 y 2 ) y = 2π r 2 y 2π y. (continue)

3 Since we are looking for the local maximum of this polynomial function for the volume of the cyliner, we can take the erivative of the function for V an set it equal to zero: y V = y (2π r 2 y 2π y ) = 2π r 2 6π y 2 = 0 6 / π y 2 = 2 / π r 2 y 2 = 1 r 2 y = r 1 We ignore the negative square root, since this is a physical measurement. x 2 = r 2 y 2 = r 2 1 r 2 = 2 r 2 x = r 2. So we fin that the imensions of the cyliner with the largest volume that can be inscribe insie the sphere of raius r has a base raius of x = r 2 = 6 an a height of h = 2y = 2 r 1 = 2 r. The maximum possible volume for the inscribe cyliner (which is what this Problem asks for) is then V = 2π x 2 2 y = 2π r 2 Since the volume of the sphere itself is of greatest volume occupies V cyl V sph = 4 / 9 / π r 4 / / π r r = 4 9 π r. V sph 9 = 1 = r = 4 π r, the inscribe cyliner of the sphere s volume (this serves as a check that the cyliner has less volume than the sphere an actually fits insie an a check that we probably solve the problem 2 V correctly). The secon erivative = 12 π y < 0 y 2, since y is positive, so we have foun the maximum value for the volume of the cyliner.

4 2) The poster, for which we shall efine the imensions length L by with W, is to have margins of one inch at the bottom an sies an of two inches at the top, which permits a printe area of A p = ( L ) ( W 2 ) ; this is the area we wish to maximize. The poster is to have a total sheet area of L W = 180 in 2, so we will be treating the function A p = LW 2L W + 6 = 180 2L W + 6 = 186 2L W. We can use the constraint equation for the total area requirement to eliminate one of the variables, say, W, to reuce our function to one with a single variable, for which we can fin the critical number: LW = 180 W = 180 L A p = 186 2L 180 = 186 2L 540 L L A p L = L 2 = 540 L 2 2 = L 2 = 2 L 2 = W = = 60 0 = 270 L = 270 = 9 0 = in. = = in. The overall imensions for the poster shoul be about 16-½ by 11 inches, leaving a printe area of 1-½ by 9 inches. Since the length L is a positive number, the secon erivative of the printe 2 A p area function is = 540 = L 2 L L 2 ( 540 L 2 ) = < 0 L L, so the area we have foun is a maximum.

5 6) A laer is intene to reach the wall of a builing, but it must clear an eightfoot tall fence that stans four feet from that wall. We wish to fin the shortest length of laer require to arrange this; although we on t prove it, it is reasonable to assume an in fact correct for optimization that the laer must rest on the top of the fence. We can work out a proportionality using similar triangles to obtain x 8 = x + 4 H = x ft., H x where H is the height above the groun where the laer touches the wall (istance not marke on the iagram). The length of the laer can then be foun using the Pythagorean Theorem: L 2 = H 2 + ( x + 4 ) 2 = x + 4 x ( x + 4 ) 2 = ( x + 4 ) x 2 It is the length L that we seek to minimize, but since L is a positive number, it will work just as well to minimize L 2 : x (L2 ) = x ( x + 4 ) x 2 = 2 ( x + 4 ) ( x + 4 )' ( x + 4 ) ' = 2 ( x + 4 ) ( x + 4 ) = 0 x 2 2 / ( x + 4 ) = 2 / ( x + 4 ) 2 64 ; x 2 x 2 x x x 2 (continue)

6 we can now ivie through by ( x + 4 ) an multiply through by x, since neither factor can be zero, to simplify this equation to x = ( x + 4 ) 64 x 2 x + 64 x = 64 x x = 256 x = ft. The foot of the shortest possible laer woul be place this far from the fence. It is then x feet away from the wall; the top of the laer is then x x feet above the groun. The length of the laer is therefore L = H 2 + ( x + 4 ) 2 = feet. 44) This is a variation of the problems about the separation between two moving people or vehicles. One of the boats travels straight south from a ock, so it is convenient to place the origin of a coorinate system at the ock an arrange the axes so that they point north-south an east-west. If we call ue north the positive y y-irection, then we can just escribe the velocity of this first boat as = 20 km. ; t hr. since it starts from the ock at 2:00 PM, which we will call time t = 0, the position of this boat can just be given as y = 0 20 t = 20 t. The secon boat is heae ue x east, which we shall call the positive x-irection, so its velocity is = +15 km.. t hr. This boat s position can be given simply in terms of x then; since it arrives at the ock ( x = 0 ) at :00 PM ( t = 1 hour ), we can fin its coorinate function from a point-slope equation: ( x 0 ) = 15 ( t 1 ) x = 15t 15.

7 This allows us to use the istance formula to express the separation between the two boats as a function of time: s 2 = (15t 15) 2 + ( 20 t ) 2. We are intereste in fining when the minimum of the function occurs, so we will look for its critical number. Because the separation s is a positive number, it will be fin to work out the minimum for s 2 : t (s2 ) = t [ (15t 15)2 + ( 20 t ) 2 ] = 2 (15t 15) (15) + 2 ( 20 t ) ( 20) = 0 / 2 (225t 225) + / 2 (400 t ) = 0 625t 225 = 0 t = = 9 25 or 0.6 hour. Thus, the boats are closest to one another at 0.6 hours = 21.6 minutes after 2:00 PM, or at 2:21:6 PM. fin If the question ha aske what the minimum separation is at that time, we woul s 2 = ( ) 2 + ( ) 2 = ( 9.6) 2 + ( 7.2) 2 = 144 km. 2 s = 12 km. The secon erivative for s 2 we ve foun is the minimum. is 2 t 2 (s2 ) = 625 > 0, so the separation 47) The pipeline that is to be run from the oil refinery to the storage tanks on the other sie of the river is built so that the overlan portion follows the river bank an then crosses beneath the river along a straight path. The river is 2 km. wier an the tanks are 6 km. ownriver from the refinery (we won t concern ourselves with the reasons this arrangement occurre ). It costs twice as much ($800,000 per km.) to run the pipeline uner the river as it oes to run it over lan ($400,000 per km.); we wish to fin the point at which the pipeline shoul begin its passage uner the river so as to minimize the total cost of construction.

8 If we place the transition point at a istance x km. ownriver from the refinery, then the portion uner the river will follow the hypotenuse of a right triangle with legs equal to the remaining ownriver istance, 6 x km., an the with of the river, 2 km. The length of the uner-river portion of the pipeline is then makes the total cost for constructing the pipeline (6 x ) km. This C = 4 x + 8 (6 x ) hunre thousan ollars ; this is the function we want to minimize. Something to keep in min is that this particular plan for the pipeline limits the omain of the cost function to 0 x 6 km. ; this issue will come up later on. The critical number is foun from C x = x 4 x + 8 (6 x )2 + 4 = (6 x ) [ (6 x )2 + 4 ]' 2 (6 x ) ( 1) = (6 x ) = 0 / 4 = / 4 2 (6 x ) (6 x ) (6 x ) (6 x ) = 1 2 (6 x ) = (6 x )2 + 4 we can multiply through by the raical, since the length of the uner-river pipe will not be zero 2 2 (6 x ) 2 = (6 x ) (6 x ) 2 = 4 (6 x ) 2 = 4 6 x = ± 4 x = 6 ± 4 km. While the quaratic function appears to leave us with two solutions, the omain of the moel cost function for the pipeline exclues a value of x larger than 6 km. So the point where the pipeline shoul begin its run uner the river lies km. ownriver from the refinery. We were not aske to calculate the minimum cost, which proves to be C (6 4.85) hunre thousan or about.79 million ollars. Here is a small emonstration of the importance of optimization: the seemingly simplest plan of just running the pipeline to the point irectly opposite the storage tanks an then straight across the river woul cost or 4.00 million ollars. C simple (6 6) hunre thousan