Applied Maximum and Minimum Problems

Transcription

1 Applied Maimum and Minimum Poblems Eample. You have 00 ft. of fence with which to fence off a ectangula plot of gound on the banks of a ive. How should you aange the fence in ode to get the lagest plot. y

2 Let the dimensions of the plot be and y with y paallel to the ive and pependicula. Let A be the aea, and P the peimete of the plot. Then we know that A= y and P= + y= 00 The fist equation gives the taget quantity (the one to be maimized o minimized) in tems of basic vaiables. The second equation is called the constaint equation, and gives a elationship among the basic vaiables. The constaint equation (o equations) of a poblem ae used to wite all the basic vaiables in tems of just one of them. Then by substituting into the equation fo the taget, we can wite the taget in tems of this single vaiable. Hee we have y = 00 and so we get the equation A= (00 ) = 00

3 The conditions of the eample show that the lagest possible value of is 100, and the smallest possible value is 0. Thus the poblem become finding the absolute maimum of the function A on the inteval [0, 100] da 00 4 d = This deivative is 0 when = 50. Thus we need to test the values 0, 50, and 100 in the fomula fo aea A= (00 ) = 00 A (0) = 0 A (50) = (50)(00 100) = 5000 A (100) = 100(00 00) = 0 The maimum occus when = 50, (y = 100), and is 5000 squae feet.

4 This suggests the following pocedue. Step 1. Daw an appopiate figue and label the quantities elevant to the poblem. Step. Find a fomula fo the quantity to be maimized o minimized. Step 3. Using the constaint equations of the poblem to eliminate vaiables, epess the quantity to be maimized o minimized as a function of one vaiable. Step 4. Find the inteval of possible values fo this vaiable fom the physical estictions in the poblem. Step 5. If applicable, use the techniques of the pevious section to obtain the maimum o minimum.

5 Eample. An open bo is to be made fom a 16 in. by 30 in. piece of cadboad by cutting out squaes of equal size fom the cones and bending up the sides. What should the size of the squaes be so that the esulting bo has a maimum volume? 16 in. 30 in.

6 Solution. Clealy the minimum value of is 0 and the maimum is 8. The volume V of the esulting bo is V= (16 )(30 ) = So we need the absolute maimum of this function on the inteval [0, 8]. dv ( ) d = + = + This is 0 at = 1 and = 10/3 (using the quadatic fomula). The citical point = 1 is outside of the inteval [0, 8], so we keep the point 10/3 and addin the end points 0 and 8. The thee values of V that esult ae espectively V (0) = 0 V (10/3) = 19600/7 76 V (8) = 0

7 Thus the maimum value occus when = 10/3.

8 Eample. A cylindical can is made to hold 1 lite of oil. Find the dimensions of the can that will minimize the cost of the metal to manufactue the can. h

9 Solution. In ode to hold a constant volume, it is clea that if the height of the can inceases, the adius of the base must decease, and convesely. Thus in pinciple, eithe the adius o height can take any value fom 0 to infinity. Then let V be the volume in cm 3 and S the suface aea in cm. V= π h= 1000 S= π + πh In this case, the fist equation is the constaint equation. Fom it we can lean that h= 1000 π Then 1000 S= π + π π 000 = + π

10 Thus the mathematical poblem becomes: Find the absolute minimum of the function in the inteval (0, ) ( 3 S () = 4π = π 500) S () = π In the inteval (0, ), this function has only the citical point = π It is clea that the deivative of S is positive fo > 0, and negative fo < 0. Thus by the fist deivative test, S has a elative minimum at 0. Thus it has an absolute minimum thee.

11 Hee is the gaph of the function S. When = 0, we have h π 0 = 0 It follows that 500 h = = 0. At the most efficient size, 0 π 0 the height and the diamete of the base ae equal, and ae appoimately cm. = Howeve, since ( ) π

12 Eample. A bo shaped wie fame consists of two identical wie squaes whose vetices ae connected by fou staight wies of equal length. If the fame is to be made fom a wie of total length L, what should the dimensions be to obtain the bo of geatest volume? y Solution. The volume is V = y, and the total length of the L wies is L = 8 + 4y. Thus we see that y= 4

13 And theefoe we need to maimize V= y= L 3 4 Subject to the estiction that 0 L/8. dv L 6 L = = 6 d This is 0 at = 0 and = L/1. If we conside the thee points 0, L/1, L/8, we see that the espective values of V ae: V (0) = 0 L 3 V L = V L 0 8 = The maimum volume then occus when = L/1. In this case we also have 4y = L-8L/1=L/3, and so y = L/1. The optimal bo is a cube whose volume is L 3 /178.

14 Eample. Show that a ight cicula cylinde of geatest volume inscibed in a ight cicula cone has volume equal to 4/9 of the volume of the cone. h y

15 By simila tiangles, we have: h y h y =, so y= h ( ) 3 h V= V = π y= π ( ) πh πh cyl = We must maimize this epession in the inteval [0, ]. dv 3 3 = πh π h = πh d This is 0 when = 0 o = 3 The volume of the cylinde is 0 at both end points, 0 and. At /3 we have

16 y= h h = 3 3 Then the volume of the cylinde is V 4 = π h π h 3 = 3 7 The volume of the cone is V = 1 πh cone 3 It follows that the atio V is V cone 4π h π = h 9

17 Eample. Find all points on the cuve closest to the point (0, ). y= 1 that ae

18 Let (, y) be any point on the cuve. The distance between (, y) and (0, ) is given by the fomula: ρ= ( 0) + ( y ) = + y 4y+ 4 Rathe than tying to minimize this function, it is clealy enough to find the places whee ρ = + y 4y+ 4 is a minimum. The constaint equation is that y= 1, so ρ = (1 + y) + y 4y+ 4= y 4y+ 5 We must theefoe find the absolute minimum of sy ( ) = y 4y+ 5 on the inteval (, ), whee s is ρ.

19 ds 4y 4 4( y 1) dy = = Hence the only citical point of the function is at y = 1, and this is a local minimum by the fist deivative test. Theefoe it is an absolute minimum. The closest points theefoe have y = 1, and =±. They ae (,1) and (,1). At these points, s = 3, so the actual minimum distance is 3.