3.6 Applied Optimization

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1 .6 Applied Optimization Section.6 Notes Page In this section we will be looking at wod poblems whee it asks us to maimize o minimize something. Fo all the poblems in this section you will be taking the deivative of something and setting it equal to zeo. EXAMPE: What is the smallest peimete possible fo a ectangle whose aea is 6 squae inches, and what ae the dimensions? Fo this one we can come up with two equations. We will use fo length and W fo width. The fist equation we have is W 6 since aea equals length times width. The second equation is fo the peimete, and this is W P. We want to get one equation with a single vaiable, so we can take the fist equation and solve fo eithe o W and put it into the second equation. I will solve the fist equation fo W: W 6. Now we 6 can put this into the second equation to eplace the W: P. This is the same as 7 P. Taking the deivative we will get: P 7. We need to set this equal to zeo and ewite this without 7 7 negative eponents: 0. Add the second tem to the othe side to get:. Coss multiplying will give us 7. You will get 6, so 6. Howeve we only want positive numbes, so we will just use = 6. We can use the equation W 6 6 to find W. We will get W 6. So = 6 in and W = 6 in. 6 To find the actual peimete, just put these numbes back into W P. So P (6 (6 4 in. EXAMPE: An open bo of maimum volume is to be made fom a squae piece of mateial, 4 inches on a side, by cutting equal squaes fom the cones and tuning up the sides (see figue. Fist wite an equation fo the volume, V, of the bo as a function of. Then find the maimum volume of the bo. To get the equation we notice that the height of this bo is. The length and width ae both 4. The fomula fo volume is V = WH. So we have V ( 4 (4 (, o V ( 4. We need to take the deivative and set it equal to zeo. Using the poduct ule we will get: V (4 ( (4 ( We can facto out 4 : V ( 4 ( 4 4 which is V ( 4 (4 6. Setting this equal to zeo you will get = 4 and =. Even though we get two answes, the value = will not be included because 4 ( = 0 which means that a bo will have a length o width of zeo which is impossible. Theefoe = 4 is the only value fo that makes sense. To find the maimum volume, put this in the ORIGINA equation fo volume, which is V ( 4. You will get V 4(4 (4 04. Theefoe the maimum aea is 04 squae inches.

2 Section.6 Notes Page EXAMPE: A ectangle is bounded by the -ais and the semicicle y 9 (see figue. What length and width should the ectangle have so that its aea is a maimum? What is the maimum aea? A A (9 9 9 ( (9 Fist we need to find the aea of the ectangle. The length is going to be since thee ae two s in the figue. The height of the bo depends on whee it hits the semicicle. The height will be equal to y, so y 9. So now we can find the aea fomula: A 9. We need to find this deivative and set it equal to zeo. This will involve the poduct ule. Fist I will ewite the poblem as: A (9. Now take the deivative: ( Poduct and chain ules used hee. Now simplify. 9 Now set the deivative equal to zeo. Move one tem to the left side solve fo. You will have 4 8 Then. Now coss multiply to get (9. Distibuting will give, so 9 8. Now. When you take the squae oot, just take the positive oot.. We know the length of the ectangle is, so the length =. The height of the ectangle is y 9. We will put ou answe fo into this fomula: y y 9. The maimum aea is these two answes multiplied togethe: A 9.. So EXAMPE: Two sides of a tiangle have lengths a and b, and the angle between them is. What value of will maimize the tiangle s aea? (Hint: A ab sin In this case a and b ae to be teated as constants, o odinay numbes. To find the maimum aea we need to take the deivative and set it equal to zeo. We will take the deivative with espect to. No poduct ule is needed hee because the only vaiable is : A ab cos. We now set this equal to zeo: 0 ab cos. Now solve fo cosine: 0 cos. To solve fo we need to take the invese cosine of both sides:. cos 0. We will get

3 Section.6 Notes Page EXAMPE: A anche has 00 feet of fencing which to enclose two adjacent coals. What dimensions should be used so that the enclosed aea will be a maimum? (See figue y In this poblem I labeled the width and the length y. It doesn t matte which vaiable you use. I can ceate two equations with this one. The fist equation is y 00. This deals with the peimete. The second equation is A y. You want to get one equation with a single vaiable, so we can solve fo eithe y o y. Solving fo y we get: y 00. You want to put this into A y fo y to get A 00 which is A 00. The deivative is: A 00. Setting it equal to zeo we will get 0 00, so. To find the y, we can use the equation y 00 and put in fo 00. You will get: y which simplifies to y = 50. So the dimensions ae by 50. EXAMPE: You ae designing a poste to contain 50 squae inches of pinting with a 4 inch magin at the top and bottom and a inch magin at each side. What oveall dimensions will minimize the amount of pape used? Fist we should daw a pictue and label sides. I will make be the height of the pinted mateial and y to epesent the? in the pictue below. Fom this pictue we can make two equations. Fist, the aea of the pinted mateial is 50 squae inches, so 50 = y. Anothe equation will involve the entie piece of pape including the magins. Hee is anothe equation: A = ( + 4(y + 8. We need to solve the fist equation fo eithe o y and 50 substitute it into the second equation. I will solve fo y in the fist equation: y. 50 Now substitute into the second equation: A ( 4 8. Instead of applying the poduct ule it may be easie to distibute fist: A This 00 simplifies to A 8 8. It is now time to take the deivative. We will ewite A as: 00 A The deivative is: A Now set it equal to zeo: 0 8. So Coss multiply: Divide to get: 5, so 5. We can t have a negative side, 50 so = 5 in. This means that y 0 in. Howeve the question asks us the find the oveall dimensions, so 5 the height will be = 8 inches and the width is = 9 inches. Theefoe the oveall dimensions of the pape ae 9 inches by 8 inches.

4 Section.6 Notes Page 4 EXAMPE: A 08 cubic foot bo with a squae base and an open top is to be constucted fom sheet metal of a given thickness. Find the dimensions of the tank with minimum weight. A bo with the minimum weight means we need to minimize the suface aea. To find the suface aea, we need to find the aea of each individual side. Thee is one bottom that is by. Then thee ae fou sides that have an aea of time y. So we y have S 4y. The volume of the bo would be V y. Since we ae give the volume, we have the fomula 08 y. We will solve this equation fo y and 08 substitute it into the othe equation. y. Now put this into the suface aea equation to get: S 4. This can be simplified to: S 4. Now we take the deivative: S 4. Setting this equal to zeo we get: 4 4 0, so. Cleaing the faction you will get 4, so 6 08 and = 6. Then we can put 6 in fo in y to get y =. So ou dimensions ae,6,6. EXAMPE: You have been asked to design a 000 cubic centimete can shaped like a ight cicula cylinde. What dimensions will use the least mateial. NOTE: The suface aea fomula fo a ight cicula cylinde is S h. The aea of a ight cicula cylinde is A h. We will ignoe the thickness of the mateial and the waste in manufactuing. Fist we ae given that the volume should be 000 cubic centimetes, so one equation we will have is 000 h. We want to eliminate one vaiable in the suface aea equation, so in ou volume equation we can solve fo eithe o h. It will be easie 000 to solve fo h since that avoids squae oots: h. Now we will substitute this into the suface aea fomula: S. This simplifies to S. Because we want to minimize the amount of mateial used, this equies us to take the deivative and set it equal to zeo. Fist we will ewite S as: S Now we take the deivative using the powe ule: S This can be 4000 ewitten as: S 4. We now want to set it equal to zeo and solve fo : , 4, , 000 0, so ou answe fo is: 6. 8 cm. Then 000 we also want to find h: h, so 000 h. This simplifies to h cm. So this means 0 we have a special elationship. We see that the height needs to be twice the adius in ode to use the least mateial.

5 Section.6 Notes Page 5 EXAMPE: The height above gound of an object moving vetically is given by s 6t 96t with s in feet and t in seconds. Find a. the object s velocity when t = 0, b. its maimum height and when it occus, and c. its velocity when s = 0. a. The deivative of position will give velocity, so s v t 96. Now we put in a zeo fo t: v (0 96. So the velocity is 96 ft/sec. b. We will take ou velocity and set it equal to zeo. This will occu when the object is at its maimum height: 0 t 96. Solving fo t will give us t = seconds. Now we put this into the oiginal equation fo s: s 6( 96(. So s = 56 feet. This is the height. c. When s = 0 we have 0 6t 96t. Factoing this will give: 0 6( t 6t 7, o 0 6( t ( t 7. Solving fo t gives t = - and t = 7. We will ignoe the negative time, so t = 7. Now that we know the time, we put this into the velocity fomula: v (7 96 = -8 ft/sec, which means it is falling back down.

Review Exercise Set 16

Review Exercise Set 16 Review Execise Set 16 Execise 1: A ectangula plot of famland will be bounded on one side by a ive and on the othe thee sides by a fence. If the fame only has 600 feet of fence, what is the lagest aea that