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Section 3.3B Word Problems: Maximizing (Finding the Maximum using the Vertex) Example1: At a concert, organizers are roping off a rectangular area for sound equipment. There is 160m of fencing available to create the perimeter. What dimensions will give the maximum area, and what is the maximum area? Steps: 1) Write an equation for the perimeter, and write an equation for the area for the rectangle. 2) Use the two equations to create a quadratic function in standard form. 3) Change the quadratic function into vertex-graphing form. 4) Identify the maximum area, and then the dimensions for the maximum area. y 1) If we let x be the width and y be the length, A is the area. Perimeter = 2x + 2y with x and y being the width and length So 160 = 2x + 2y (Solve for y to create a function.) 160 – 2x = 2y 80 – x = y y = 80 – x x Sound Equipment y Area = xy A = xy 2) Now combine the 2 equation from part 1) to find an equation to maximize the area. y = 80 – x A = xy A = x(80 – x) A = 80x – x2 A = -x2 + 80x b formula. 2a Since all we need is the vertex, the simpler and quicker way is to use the formula to find the vertex. A = -x2 + 80x a = -1, b = 80 b 80 (x, A) p=− =− = 40 So the vertex is (40, ??) 2a 2(−1) Substitute 40 in for x and we get: A = -(40)2 + 80(40) = -1600 + 3200 = 1600 Our vertex is (40, 1600) 3) Now complete the square to change into vertex-graphing form OR use p = − 4) Using the vertex, our maximum area is 1600m2 and one dimension of the box is 40m. Since A = xy 1600=40y so y = 40 The box is 40m by 40m. x Graph of the equation: A = -x2 + 80x A 1800 1600 1400 1200 1000 800 600 400 200 x 0 10 20 30 40 50 60 70 80 90 100 Example2: A sporting goods store sells basketball shorts for $8. At this price their weekly sales are approximately 100 items. Research says that for every $2 increase in price, the manager can expect the store to sell five fewer pairs of shorts. Determine the maximum revenue the manager can expect based on these estimates. What selling price will give that maximum revenue, and how many shorts will be sold? Let x = # of $2 price increases Let R = the revenue Based on no changes, the revenue would have been R = (price)(shorts_sold) = (8)(100). With the expected changes: The new price = 8 + 2x {for every increase, price goes up by $2} The shorts sold = 100 – 5x {for every increase, number of shorts sold goes down by 5} Formula: R = (8 +2x)(100 – 5x) To get the maximum revenue, rewrite the above as a quadratic function and find the vertex (maximum values). R = 800 – 40x + 200x – 10x2 R = -10x2 + 160x + 800 Use the formula p = − b to help find the vertex. 2a R = -10x2 + 160x + 800 a = -10, b = 160 p=− b 160 =− = 8 So the vertex is (8, ???) 2a 2(−10) Substitute 8 in for x in the equation R = -10x2 + 160x + 800 to find the other part of the vertex. R = -10(8)2 + 160(8) + 800 R = -640 + 1280 +800 R = 1440 So the vertex is (8, 1440) This means there are 8 increases of $2 with a maximum revenue of $1440. The new price will be 8 + 2x = 8 + 2(8) = $24 The number of shorts sold will be 100 – 5x = 100 – 5(8) = 60 shorts Graph: R 1600 1400 1200 1000 800 600 400 200 0 -24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24 x Word Problem questions for students to try: #1. Cathy has 48m of fencing to make a rectangular pen for her pet. What is the maximum area? Answer: 144m2 Let x = width Let 24-x = length (P = 2L + 2W 48 = 2L + 2W 24 = L + W L = 24 – W) A = x(24-x) A = 24x – x2 A = -x2 + 24x Find vertex (p, q) to find maximum Area. b 24 p= − = − = 12 2a 2(−1) q = -(12)2 + 24(12) q = -144 + 288 q = 144 So the maximum area = 144m2 #2. The Summer Theatre charges $10 per ticket and it has had a full house of 500 nightly. The manager estimates that the ticket sales would decrease by 50 for each $2 increase in ticket cost. What is the most profitable price to charge for each ticket? At this new price what is the maximum revenue the Summer Theater can expect to generate? Answer: ticket price = $15 max revenue = $5625 Let x = number of $2 increases Let R = revenue R = price x numberoftickets R = (10 + 2x)(500 – 50x) R = 5000 – 500x + 1000x – 100x2 R = -100x2 + 500x + 5000 Find vertex (p, q) to find the Max revenue. b 500 p= − = − = 2.5 2a 2(−100) q = -100(2.5)2 + 500(2.5) + 5000 q = 5625 The Max revenue is $5625 . The price is 10 + 2(2.5) = $15 per ticket #3. A management firm has determined that 60 apartments in a complex can be rented if the monthly rent is $900, and that for each $50 increase in the rent, three tenants are lost with little chance of being replaced. What rent should be charged to maximize revenue? What is the maximum revenue? Answer: $950 per month for rent max revenue = $54,150 Let x = number of $50 increases Let R = revenue R = monthly rent x numberofapartment R = (900 + 50x)(60 – 3x) R = 54000 – 2700x + 3000x – 150x2 R = -150x2 + 300x + 54000 b 300 p= − = − = 1 2a 2(−150) q = -150(1)2 + 300(1) + 54000 q = 54150 The Max revenue is $54150 . The monthly rent is 900 + 50(1) = $950 per apartment Textbook Assignment: Page 194-196 #14, 18, 19