HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Hawkes Learning Systems: College Algebra Section 4.2b: Maximization/Minimization Applications of Quadratic Functions HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Objective o Interlude: maximization/minimization problems. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Maximization/Minimization Problems Many applications of mathematics involve determining the value (or values) of the variable x that returns either the maximum possible value or the minimum possible value of some function f(x). Such problems are called max/min problems for short. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Maximization/Minimization Problems We are now able to solve max/min problems involving quadratic functions. If the parabola opens upward, we know that the vertex is the lowest (minimum) point on the graph. If the parabola opens downward, we know that the vertex is the highest (maximum) point on the graph. Either way, we locate the vertex by completing the square. HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Maximization/Minimization Problems We can shorten the process of locating the vertex by completing the square on the generic quadratic: f x ax 2 bx c As always, we begin by factoring 2 b the leading coefficient a from the a x x c a first two terms. To complete the square, we add 2 b b2 b2 b the square of half of inside the a x x 2 c a a 4 a 4 a parentheses. This means we also b 4ac b 2 a x 2a 4a b 4ac b 2 Vertex: , 2 a 4 a 2 have to subtract a times this quantity outside the parentheses. Note: the vertex must lie at b b , f 2a 2a . HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 1: Maximization/Minimization Problems Find the vertex of the following functions. 2 f x 3 x x7 a. 1 4 3 7 1 2 1 85 , vertex , 4 3 23 6 12 Remember that the vertex lies at b 4ac b 2 , . 2 a 4 a b. f x 5 x 2 2 x 4 2 2 4 5 4 2 1 19 , vertex , 4 5 2 5 5 5 Continued on the next slide… HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 1: Maximization/Minimization Problems (cont.) Find the vertex of the following function. 2 f x x 3x 6 c. 2 3 4 1 6 3 3 33 , vertex , 4 1 2 1 2 4 HAWKES LEARNING SYSTEMS Copyright © 2011 Hawkes Learning Systems. All rights reserved. math courseware specialists Example 2: Maximization/Minimization Problems A farmer plans to use 2500 feet of spare fencing material to form a rectangular area for cows to graze against the side of a long barn, using the barn as one side of the rectangular area. How should he split up the fencing among the other three sides in order to maximize the rectangular area? MOO! x x 2500 2x If we let x represent the length of one side of the rectangular area then the dimensions of the rectangular area are x feet by 2500-2x feet (see image above). We will let A be the name of our function that we wish to maximize in this problem, so we want to find the maximum possible value of A x x 2500 2 x . Continued on the next slide… HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 2: Maximization/Minimization Problems (Cont.) Note: If we multiply out the formula for A, we get a quadratic function. A x x 2500 2 x 2 x 2 2500 x We know this function is a parabola facing down. We also know that the vertex is the maximum point on this graph. Remember, the vertex is the point b 4ac b2 b b 2a , 4a or 2a , f 2a . So, plugging in the values, we get the vertex 625, A 625 . Therefore, the maximum possible area is A(625): A(625) 781250 square feet.

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