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Pre – Calc Lesson 2.4 Finding Maximums and Minimums of Polynomial Functions

Pre – Calc Lesson 2.4 Finding Maximums and Minimums of Polynomial Functions For quadratic functions: f(x) = ax 2 + bx + c To fin d the max. or min. 1 st : find x = ??

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Pre – Calc Lesson 2.4 Finding Maximums and Minimums of Polynomial Functions

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  1. Pre – Calc Lesson 2.4 Finding Maximums and Minimums of Polynomial Functions For quadratic functions: f(x) = ax2 + bx + c To fin d the max. or min. 1st : find x = ?? Hint: axis of symmetry: Then substituting that value back in to the ‘rule’ for ‘x’, you can find the ‘y’ value that goes with that ‘x’ value. This ‘y’ value is your ‘Max’ or ‘Min’ Example 1: A rectangular dog pen is constructed using a barn wall as one side and 60m of fencing for the other three sides. Find the dimensions of the pen that give the greatest possible area.

  2. Draw a picture -- see page 69 at the top • Let ‘x’ = length (in meters) of one on the sides • perpendicular to the barn. • Determine the length of the other sides in terms • of ‘x’. • length = x • width = ?? 60 – 2x see diagram • Recall that the area of a rectangle : A = ?? • length x width • Therefore A(x) = x( ?? ) =x(60-2x) • A(x) = ???? = 60x – 2x2 • So if x = - b - ??- 60 - ??  - 60 = 15 • 2a ?? 2(-2)?? -4

  3. Therefore the maximum occurs when x = ?? 15 (Note: this is not the maximum, just where it occurs) So the dimensions would have to be ??????? 15 by (60-2(15) ) or 15 by 30 So the maximum area to enclose would be: ?????? 450 - this could be obtained either by formula or plug 15 back in to original equation Example 2 Squares with sides of length ‘x’ are cut from the corners of a rectangular piece of sheet metal with dimensions 6” x 10” The metal is folded to make an open-top box. What is the maximum volume of such a box? 1st: Draw a picture see example 2 at bottom of page 69. 2nd: Identify the dimensions in terms of ‘x’

  4. height = x length = 10 – 2x width = 6 – 2x 3rd: Come up with a formula. Recall the volume of a rectangular solid: V = l x w x h Therefore: V(x) = (10-2x)(6-2x)(x) Now leaving this in factored form: take: 10- 2x = 0 and 6 – 2x = 0 and x = 0 10 = 2x 6 = 2x x = 0 5 = x 3 = x x = 0 Plot these points on a graph, find the: Leading term: 4x3 a = 4  (graph rises) Constant term: 0  y-intercept

  5. Now make a sketch -- and rationalize that the local max. Occurs between the roots ‘0’ and ‘3’. Now use your grapher to find the local max is: y = 32.84 when x = 1.21 Therefore to answer the question: The maximum volume is 32.84 in3 (Hw: CE #1-4 WE: #1,3,9,10,11

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