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1 a. H y) - /4x -to izx + x'/e ^W-^ox^-^^x^+ix^2 -w X n -?^X G 4- zx'/2 b- ax)~ X lk?x ryv)- InV + xgf ) -- Inx -+1 ca^=e5)c^+f7^0 U y.- 5e5y/xVnv-e)+e5Y^3-+n) d' #*) =[Kx^-i)]^] Hx)-- X3f2x-> (V-# +n^x3+ev-i)k2(v--5)

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38 Solutions to optimization problems Found a mistake? Please to grigory.bonik@uconn.edu. Each solution will follow these six steps: 1. Define variables; 2. Write equations relating variables to each other; 3. Express the quantity that we want to optimize as a function of a single variable; 4. Find a resonable interval for the variable; 5. Find the absolute extremum of the function on the interval; 6. Compute the final answer. Problem 29. You want to paint a picture. You have enough wood to make a frame that is 400 cm around. What s the largest picture you can make? (largest means of largest possible area ). Solution. 1) First, we need to clearly define our variables. Let x and y be the dimensions of the pciture in centrimeters, as shown on the picture below. x Let P be the perimeter and A be the area. 2) Next, we have to establish some equations. We have enough wood to make a frame that is 400 cm around, which means that the perimeter must be 400 cm, i.e. P =400. On the other hand, perimeter is the total length of all sides: Hence we obtain the following equation: P = x + x + y + y =2(x + y). y 2(x + y) = 400 Next, using the formula for rectangle area, we write x + y = 200 (1) A = xy. (2) 1

39 3) Since our goal is to maximize the area, we need to express it as a function of a single variable (either x or y, not both!). To accomplish this, let s solve equation (1) for y: Plug in the latter into equation (2): y =200 x. A = x (200 x) =200x x 2. To indicate that the area A is a function of x, let swrite A(x) =200x x 2. (3) 4) Now we have a function to maximize. But before looking for the absolute maximum, we have to come up with a reasonable interval for x. Itmakesnosenseforx to be negative, because it s a length in centimeters. However, it makes certain sense to allow x to be equal to zero, so we want x 0. We cannot make x arbitrary big, because we have only limited amount of wood. To find the biggest possible value for x, imagine that we spend all wood on the vertical sides and no wood on the horizontal sides, i.e. y =0. Thenfromequation(1) weconcludethat200is the biggest possible value for x. Hence our interval is [0, 200]. 5) Now we have a function A(x) definedbyequation(3) andtheinterval[0, 200]. We can now apply our standard process of finding the absolute extrema. Take the derivative: A (x) =200 2x. Set it equal to zero and solve for x to get critical points: 200 2x = 0 2x = 200 x = 100. The only critical point is x = 100. Now test the critical point and the endpoints (plug in to A(x), not the derivative!): A(0) = =0, A(100) = =10000, A(200) = = 0. We see that that A(100) is the biggest value and therefore the absolute maximum of A(x) on the interval [0, 200]. 6) We conclude that the biggest picture we can make has the area cm 2. To find its dimensions, we need to find y: y =200 x = = 100. Therefore the biggest picture is a square of size 100 cm 100 cm, which is predictable. 2

40 Problem 30. A fence is to be built around a 300 square foot rectangular field. One side costs twice as much per unit length as the other three. Find the dimensions of the enclosure that minimizes total cost. Solution. 1) Let x and y be the dimensions of the field in feet, as shown on the picture below. x N y Let C be the total cost, and A be the area. 2) We know that one side costs twice as much per unit length as the other three. Assume it s the north wall. We don t know what exactly the price per unit length in dollars is, but the simple logic suggests that the answer doesn t depend on that. So we can assume that the north wall costs $2 per foot and other three walls cost $1 per foot. So the total cost is C = 2y + 1x +1x + 1y =3y +2x. (4) north east and west south Now let s write the formula for the area of a rectangle: A = xy. By the problem statement, we want the area to be 300 ft 2 : xy =300. (5) 3) We want to minimize the cost, so we need it as a function of one variable. Solve (5) for y: y = 300 x. (6) Then plug it in to the equation for the cost: C =3y +2x =3 300 x +2x = 900 x +2x, so C(x) = 900 x +2x. 4) We need a reasonable interval for x. Ofcourse,x cannot be negative, since it s a length in feet. We can make it as small or as big as we want, because we have no restrictions on the perimeter. But we cannot make it equal to zero, because it would make the area zero. So the interval for x is (0, + ). 3

41 5) Now we have a function and an interval. The interval, however, is not closed, which means we cannot apply our standard procedure. Anyway, we need to find the critical points. Take the derivative of C(x) and set it equal to zero: Afractioniszerowhenthetopiszero,so C (x) = 900 x 2 +2, 0 = 900 x +2, x2 0 =. x x 2 = 0, 2x 2 = 900, x 2 = 450, x = ± 450. There are two critical points, 450 and 450. However, 450 is outside of our interval (0, + ), therefore we don t need to consider it. We need to justify that C(x) indeedhasanabsoluteminimumat 450. Since it s the only critical point in the interval, it suffices to show that it is a relative minimum. We need to test the sign of the derivative to the left and to the right of our critical point: C (1) = < 0, 12 C (100) = > f (x) f(x) We see that C(x) indeed has a relative minimum at x = 450. Since it s the only critical point on the interval, it must be the absolute minimum. 6) We are asked to find the dimensions which minimize the cost. We already have one of them, x = 450 = To find y, useequation(6): y = =10 2. So, the dimensions that minimize the cost are 15 2ft 10 2ft. 4

42 Problem 31. The manager of a large apartment complex knows from experience that 100 units will be occupied if the rent is 296 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 4 dollar increase in rent. Similarly, one additional unit will be occupied for each 4 dollar decrease in rent. What rent should the manager charge to maximize revenue? Solution. 1) Let x be the number of 4 dollar increases (when x is negative, then x is the number of 4 dollar decreases). Let p be the rent in dollars and n the number of occupied units. Let R be the revenue. 2) Using our definition of p and x we can immediately write By the problem statement, p =296+4x. n =110 x. The revenue is the number of occupied units times the rent: R = np. 3) We want to maximize the revenue, so let s express it in terms of single variable x: R(x) =(110 x)( x). 4) Let s find a reasonable interval for x. We cannot make x larger than 110, because n would then become negative. On the other hand, we cannot make it smaller than 296 = 4 74, because that would make the price negative. Therefore the interval is [ 74, 110]. 5) We need to find the absolute maximum of R(x) ontheinterval[ 74, 110]. Take the derivative: R (x) = 1 ( x)+(110 x) 4= 8x 144. Set it equal to zero to find the critical points: 8x +144 = 0 8x = 144 x = 18 Plug in the critical point and the endpoints: R( 74) = 0, R(18) = something big, R(110) = 0. Hence R(x) hastheabsolutemaximumatx =18. 6) To find the optimal price, recall that p =296+4x = = 368. Hence the optimal price is $368. 5

43 Problem 32. A deli sells 320 sandwiches per day at a price of 4 each. A market survey shows that for every 0.10 reduction in the price, 20 more sandwiches will be sold. How much should the deli charge in order to maximize the revenue? Solution. 1) Let x be the number of 0.10 reductions, p be the price, n the number of sandwiches sold, and let R be the revenue. 2) Equations: 3) Revenue in terms of single variable: p = x, n = x, R = np. R(x) =(4 0.10x)( x) 4) In this problem, x should be non-negative, because we can only reduce the price, not increase it. It cannot be bigger than 40, because this would make the price negative. Hence the interval is [0, 40]. 5) We need the absolute maximum of R(x) on[0, 40]. Find the derivative: The only critical point is x =12. Plugin: R (x) =48 4x. R(0) = 1280, R(12) = 1568, R(110) = 0. Hence R(x) hastheabsolutemaximumatx =12. 6) The optimal price is p = =

44 Problem 33. A straight piece of wire 8 feet long is bent into the shape of an L. What is the shortest possible distance between the ends? Solution. 1) Let x and y be lengths of the sides of L in feet, as shown on the picture. Let d be the distance between the ends. x d 2) Since the total length of the wire is 8 feet, we can write y x + y =8. By the Pythagorean theorem, d = x 2 + y 2. 3) Express d as a function of x: d(x) = x 2 +(8 x) 2 = 2x 2 16x +64. Note that the square root is smallest when the number inside it is smallest (as long as it stays non-negative, of course). This means that instead of d(x), we can minimize f(x) =2x 2 16x ) Obviously, x can go from 0 to 8 feet, i.e. the interval is [0, 8]. 5) Derivative: f (x) =4x 16. The only critical point is x =4. Plugin: f(0) = 64, f(4) = 32, f(8) = 64. Hence f(x) (andd(x) aswell)hastheabsoluteminimumatx =4. 6) The shortest distance between the ends is d(4) = = 32. 7

45 Problem 34. You are opening a kennel, and creating a rectangular area with room for 5 dogs, but the dogs must be kept separate from each other. The outside fencing costs $10 per foot, and inside barriers to separate dogs cost half as much. You have $2000 to spend what s the largest area you can enclose? (You want to separate the dogs by dividing the area using parralel barriers.) Solution. 1) Let x and y be dimensions of the area in feet, as shown on the picture. Let C be the total cost and A be the total area. x y 2) The total cost is the cost of the exterior fence plus the cost of the interior barriers: C =10(2x +2y) exterior fence + 4 #of barriers 5x cost of one barrier =40x +20y. (7) We want to spend $2000: Next, the total area is 40x +20y = 2000, Here we neglect the thickness of the interior barriers. 3) Solve (8) fory: y =100 2x. Plug in to (9) togettheareaasafunctionofx: 2x + y = 100. (8) A = xy. (9) A(x) =x(100 2x) =100x 2x 2. 4) Since x stands for length in feet, it cannot be negative. However, it makes certain sense to allow it to be zero. To find the biggest value of x that makes sense, imagine that we spend all money on vertical walls, i.e. we make y equal to zero. Then using (8) wecan write 40x =2000, i.e. x = 50. Hence the interval is [0, 50]. 5) We need to find the absolute maximum of A(x) on[0, 50]. Take derivative: A (x) =100 4x. 8

46 Find critical points: 100 4x = 0, x = 25. The only critical point is x =25. Timetoplugin: A(0) = 0, A(25) = 1250, A(50) = 0. We conclude that A(x) hastheabsolutemaximumatx =25. 6) The largest area is 1250 square feet. Good luck on the exam! 9