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Max and min problems

maximum. vertex. minimum. Max and min problems. As we saw previously the vertex of a parabola represents the: maximum if the parabola is sad, a < 0. minimum if the parabola is happy, a > 0. Example 1.

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Max and min problems

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  1. maximum vertex minimum Max and min problems • As we saw previously the vertex of a parabola represents the: • maximum if the parabola is sad, a < 0. • minimum if the parabola is happy, a > 0.

  2. Example 1 • Tania’s Toasters has found the cost of making its electric toasters depends on the number of toasters made per day. The cost (C dollars) per toaster can be calculated using the formula • C = 2x2 12x + 35 • where x is the number of toasters (in hundreds) made per day. • Graph C = 2x2 12x + 35 for x between 0 and 8. • When 200 toasters are produced in a day, what is the cost per toaster? • Last Friday the toasters cost $25 to produce, how many toasters did Tania produce? • On Monday it cost $40 to produce a toaster. How many more toasters did they produce on Monday than Friday? • At what rate per day should Tania produce toasters to keep the production costs to a minimum? • What is the minimum production cost?

  3. Example 2 (con’t) C = 2x2 12x + 35 35 25 19 17 19 25 35 49 67 b) c) d) e) f) $19 100 or 500 toasters 640  100 = 540, or 640  500 = 140 300 toasters $17

  4. shed chicken run Example 2 Farmer Sarah has 22m of chicken wire and plans to build a chicken run using an existing wall of a shed as one side of the run. x x 22  2x a) Explain why the area (A) of the chicken run can be modelled by A = 22x 2x2 where x is the width of the chicken run. b) Graph A = 22x 2x2 for values of x from 0 to ____ c) What is the maximum area of the chicken run. d) What length and width of the run will give the maximum area? a) Let the side width be x length = 22  x  x = 22  2x A = LB A = A = 11 as if x is larger than 11 the sides won’t meet. Axis of symmetry b) The largest x can be is x× (22  2x) 22x 2x2

  5. Example 3 0 20 36 48 56 60 60·5 60 56 48 36 20 0 A = 22x 2x2 c) d) Maximum area is at the vertex max area = 60·5m2 width = 5·5m length= 22  2x = 22  2 × 5·5 = 11m

  6. Today’s work Exercise 9-02 Page 337 → 339 Q2, 3, 4, 5, 8 & 10

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