OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS BC CALCULUS

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1 1 OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS BC CALCULUS 1. Read each problem slowly and carefully. Read the problem at least three times before trying to solve it. Sometimes words can be ambiguous. It is imperative to know exactly what the problem is asking. If you misread the problem or hurry through it, you have NO chance of solving it correctly.. If appropriate, draw a sketch or diagram of the problem to be solved. Pictures are a great help in organizing and sorting out your thoughts. 3. Define variables to be used and carefully label your picture or diagram with these variables. This step is very important because it leads directly or indirectly to the creation of mathematical equations. 4. Write down all equations which are related to your problem or diagram. Experience will show you that MOST optimization problems will begin with two equations. One equation is a "constraint" equation and the other is the "optimization" equation the one you are asked to maximize or minimize. The "constraint" equation is used to solve for one of the variables. This is then substituted into the "optimization" equation before differentiation occurs. Some problems may have NO constraint equation. Some problems may have two or more constraint equations. 5. Before differentiating, make sure that the optimization equation is a function of only one variable. Then differentiate using the well-known rules of differentiation. Example 1: Two positive numbers have a sum of 60. What is the maximum product of one number times the square of the second number? a + b = 60 P = a b P max =? P = a b = a (60 a) = 60a a 3 dp da = 10a 3a = 0 a(40 a) = 0 a = 0 P = 0 minimum a = 40 b = 0 P max = 3000 Example : We need to enclose a portion of a field with a rectangular fence. We have 500 feet of fencing material and a building is on one of the longer sides of the field and so won t need any fencing. Determine the dimensions of the field that will enclose the largest area. a + b = 500 A = ab A max =? A = ab = (500 b)b = 500b b = 500 4b = 0 b = 15 a = 50 da A max = 3150

2 Example 3: We need to enclose a portion of a field with a rectangular fence that has two pens of equal area and equal size. We want to enclose at total of 10,000 square feet. Determine the least amount of fencing we will need. A = ab = P = 3b + 4a P min =? P = 3b + 4a = 15000/a + 4a dp da = a + 4 = 0 a = b = P min = feet Example 4: We need to enclose a portion of a field with a rectangular fence. A river is on one side of the fence, so we won t need to fence along the river. We would like to enclose 10,000 square feet. Because of the added construction costs of building perpendicular to the river, it costs $8 a liner foot to buy and build the fence. The portion of the fence parallel to the river only cost $5 per liner foot to buy and build. What is the least amount of money we will need to construct our fence? ab = M = 8b + 5a M min =? M = 16b + 5a = 16b /b dm = 16 db b = 0 b = a = M min = $ Example 5: We want to construct a box with a square base and we only have 10 square meters of material to use in construction of the box. Assuming that all the material is used in the construction process determine the maximum volume that the box can have. a + 4ab = 10 V = a b V max =? V = a b = a 5 a a = 5a a3 / dv da = 5 3a = 0 a = 5 b = V max = a 3 =

3 Example 6: A cylindrical aluminum soda can is to hold 1 ounces of tasty beverage. What are the dimensions of the can that will minimize the amount of aluminum used? Hint: 1 US fluid ounces equals cubic inches. If the dimensions of the actual can are.6 inch diameter and 4.75 inches tall, does the actual can have the optimal dimensions? (πr )h = A = (πr)h + (πr ) A min =? 3 STRANGE??? A = (πr)h + (πr ) = πr πr + πr = πr r da = r + 4πr r = 1.511" h = " A min = inches 3 Sometimes we also want to maximize CLOSENESS! Example 7: Determine the point(s) on y = x + 1 that are closest to (0,). Points (x, y)and (0,) y = x + 1 d = (x 0) + (x + 1 ) d min =? d = x 4 x + 1 dd dx = 4x 3 x x 4 x + 1 = 0 x(x 1) x = 0 x = ± for x = 0, d = 1 d min = 3 for x = ± 3, d = for, 3 and, 3

4 4 At this point, you might be tempted to think that the optimal value is always at the critical value. HA HA HA Example 8: A feet piece of wire is cut into two pieces and once piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire cut so that the total area enclosed by both is a (a) minimum and (b) maximum? (Note: it is not required to build both shapes). Let s cut the wire into two pieces. The first piece will have length x which we ll bend into a square and each side will have length x. The second piece will then have length ( x)(constraint) and we ll bend this into 4 an equilateral triangle and each side will have length 1 ( x). 3 A(x) = x x ( x) 3 ( x) = ( x) A max =? A min =? dx = x ()( x)( 1) = x x 18 9 = 0 x = = Now, let s notice that the problem statement asked for both the minimum and maximum enclosed area and we got a single critical point. BUT: Let s notice that x must be in the range 0 x and since the area function is continuous we use the basic process for finding absolute extrema of a function. A(0) = A(0.8699) = A() = 0.5 So, it looks like the minimum area will arise if we take x = while the maximum area will arise if we take the whole piece of wire and bend it into a square. One of the extrema in the previous problem was at an endpoint and that will happen on occasion.

5 5 Sometimes we need to establish a pattern to write our primary equation (Examples 8 & 9). Example 9: The manager of an 80-unit apartment complex is trying to decide what rent to charge. Experience has shown that at a rent of $00, all of the units will be full. On the average, one additional unit will remain vacant for each $0 increase in rent. Find the rent to charge so as to maximize revenue. a pattern: a) x = the number of $0 increases. The rent for an apartment is x b) number of apartments rented: For each $0 increase, one less apartment will be rented. Since x is the number of 0 increases, the number of vacancies will be x, and the number of apartments rented will be 80 x the total revenue from all rented apartments: R = (80 x)(00 + 0x) = x 0x R max =? dr = x = 0 x = 35 dx Rent for an apartment would be $900 (very expensive), 45 rented and R = $40500 Example 10: A fence, 8 feet high, is parallel to the wall of a building and 1 foot from the building. What is the shortest plank that can go over the fence, from the level ground, to prop the wall? sin θ = 8 x cos θ = 1 y = 8 sin θ + 1 cos θ L = x + y 0 < θ < π dl 8 cos θ sin θ = dθ sin + θ cos θ = 0 8cos3 θ + sin 3 θ = 0 tan 3 θ = 8 tan θ = θ = 63.4 o 1.1 dl dθ + θ π

6 6 0, π is open interval, so L has local and absolute minimum only at θ = 1.1 x = tan θ = = 4 5 y = 1 + (1 tan θ) = = 5 L = x + y = 5 5ft 11.ft It s fun to inscribe shapes into other shapes that circumscribe the inscribed shape we re inscribing. Example 11: Find the volume of the largest cone that can be inscribed inside a sphere of radius 5. (h R) + r = R V = 1/3 π r h V = 1 3 π [R (h R) ]h = 1 3 π [Rh h 3 ] dv dh = 1 3 π[4rh 3h ] = 0 h = 0 (min) h = 4 R (max) 3 V max = 3 81 πr3 Example 1: An arch top window is being built whose bottom is a rectangle and the top is a semicircle. If there is 1 meters of framing materials what is the width of the window that lets in the most light? h + r + πr = 1 A = hr + 1 πr A(r) = r 6 r 1 πr + 1 πr = 1r r 1 πr 1 = 1 r(4 + π) = 0 r = dr 4 + π = d A = 4 π < 0 dr A max must have semicircle r = and rectangle (h r)