Feb 23 Extra Min Max Problems
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  • 1. Extra Quadratic Min/Max Problems and Solutions
  • 2. A cattle farmer wants to build a rectangular fenced enclosure divided into five rectangular pens as shown in the diagram. A total length of 120m of fencing material is available. Find the overall dimensions of the enclosure that will make the total area a maximum.
  • 3. Solution L w w w w w w L 120 = 6w + 2L L = 120 - 6w Area = w*L 2 L = (60 - 3w) Area = w(60-3w) Area = 60w - 3w 2
  • 4. 2 Area = 60w - 3w Area = -3{w 2 - 20w} Area = -3{w 2 - 20w + 100 -100} 2 - 20w + 100) -100} Area = -3{(w 2 - 100} Area = -3{(w - 10) Area = -3(w - 10) 2 - (-3)100 2 Area = -3(w - 10) +300
  • 5. L w w w w w w L 2 Area = -3(w - 10) +300 Area is maximum when w = 10m The maximum area is 300m 2 Area = L*w 300 = L*10 L = 30m
  • 6. An amusement park charges $8 admission and averages 2000 visitors per day. A survey shows that, for each $1 increase in the admission cost, 100 fewer people will visit the park. What admission cost gives the maximum profit?
  • 7. Solution Profit = Admission * Visitors A = 8 + 1x V = 2000 - 100x where x = number of times the price is increased P = (8+x)(2000-100x)
  • 8. P = (8+x)(2000-100x) P = 16000 - 800x + 2000x - 100x2 P = -100x 2 + 1200x + 16000 P = -100{x2 - 12x} + 16000 2 - 12x + 36) -36} +16000 P = -100{(x
  • 9. P = -100{(x 2 - 12x + 36) -36} +16000 P = -100{(x-6)2 -36} + 16000 P = -100(x-6)2 - (-100)36 + 16000 2 + 19600 P = -100(x-6)
  • 10. Profit = Admission * Visitors A = 8 + 1x V = 2000 - 100x where x = number of times the price is increased P = -100(x-6) 2 + 19600 So maximum profit is $19600 when x = 6 The admission price should be A = 8 + 6 = $14
  • 11. Determine the maximum area of a triangle if the sum of its base and its height is 13 cm.
  • 12. Solution b + h = 13 b = (13 - h) Area = 1/2 b*h Area = 1/2(13-h)(h) Area = (6.5 - .5h)(h) Area = 6.5h - .5h2
  • 13. 2 Area = 6.5h - .5h Area = -.5b 2 + 6.5h Area = -.5{h2 - 13h} Area = -.5{(h2 - 13h + 6.52) - 6.52} Area = -.5{(h-6.5) 2 - 6.52} 2 - (-.5)6.52 Area = -.5(h-6.5)
  • 14. 2 - (-.5)6.52 Area = -.5(h-6.5) 2 + 21.125 Area = -.5(h-6.5) The maximum area of the triangle is 21.125 cm2 when the base is 6.5 cm.