5 023 MAX - Min: Optimization

23
5023 MAX - Min: Optimization AP Calculus

description

5 023 MAX - Min: Optimization. AP Calculus. First Derivative Test for Max / Min TEST POINTS on either side of the critical numbers MAX :if the value changes from + to – MIN : if the value changes from – to +. Second Derivative Test for Max / Min FIND 2 nd Derivative - PowerPoint PPT Presentation

Transcript of 5 023 MAX - Min: Optimization

OPTIMIZATION PROBLEMS

5023 MAX - Min: Optimization AP Calculus

1OPEN INTERVALS:

Find the 1st Derivative and the Critical NumbersFirst Derivative Test for Max / MinTEST POINTS on either side of the critical numbersMAX:if the value changes from + to MIN: if the value changes from to +Second Derivative Test for Max / MinFIND 2nd DerivativePLUG IN the critical numberMAX: if the value is negativeMIN: if the value is positive2Example 1: Open - 1st Derivative test

VA=1 x=3, x=-1Undefined at 1 f(x)-113+--+-202VA4 max at -1 min at 3 and a VA at 1 why? Max of -2 at x=-1 because the first derivative goes from positive to negative at -1. Min of 6 at x=3 because 1st derivative goes from negative to positive at 3.A VA at x=1 because 1st derivative and the function are undefined at x=13Example 2: Open - 2nd Derivative Test

4LHE p. 186 CLOSED INTERVALS:Closed Interval Test

Find the 1st Derivative and the Critical Numbers

Plug In the Critical Numbers and the End Points into the original equationMAX: if the Largest valueMIN: if the Smallest value

EXTREME VALUE THEOREM:If f is continuous on a closed interval [a,b], then f attains an absolute maximum f(c) and an absolute minimum f(d) at some points c and d in [a,b]localCritical #6CLOSED INTERVALS:

Find the 1st Derivative and the Critical Numbers Closed Interval TestPlug In the Critical Numbers and the End Points into the original equationMAX: if the Largest valueMIN: if the Smallest value7Example : Closed Interval Test

Consider endpts.Absolute max absolute min8LHE p. 169 even numbersOPTIMIZATION PROBLEMSUsed to determine Maximum and Minimum Values i.e. maximum profit,least cost,greatest strength,least distance13METHOD: Set-UpMake a sketch.

Assign variables to all given and to find quantities.

Write a STATEMENT and PRIMARY (generic) equation to be maximized or minimized.

PERSONALIZE the equation with the given information. Get the equation as a function of one variable. < This may involve a SECONDARY equation.>

Find the Derivative and perform one the tests.

14Relative Maximum and Minimum FIRST DERIVATIVE TEST FOR MAX/MIN.If f is continuous on [a, b] containing a critical number,c, and differentiable on (a,b) except possibly at c and if:f / (x) changes signs from (+) to (-) then f(c) is a relative Maximum.f / (x) changes signs from (-) to (+) then f(c) is a relative Minimum.f / (x) has no sign change then f(c) is neither.

Note: c is the location. f (c) is the Max or Min value.DEFN: Relative Extrema are the highest or lowest points in an interval.Where is xWhat is y15Concavity and Max / MinSECOND DERIVATIVE TEST FOR MAX/MIN. If f is continuous on [a, b] and twice differentiable on (a,b) and if f (c) is a critical number, then if f // (c) > 0 then f (x) is concave up and c is a minimum

if f // (c) < 0 then f (x) is concave down and c is a maximum

if f // (c) = 0 then the test is inconclusive.

161 ILLUSTRATION : (with method)A landowner wishes to enclose a rectangular field that borders a river. He had 2000 meters of fencing and does not plan to fence the side adjacent to the river. What should the lengths of the sides be to maximize the area?Statement:Generic formula:Personalized formula:Which Test?Figure:yxMax areaA=lwA=xy x+2y=2000 x=2000-2y x=1000 y=500To maximize the area the lengths of the sides should be 500 meters.17Example 2:Design an open box with the MAXIMUM VOLUME that has a square bottom and surface area of 108 square inches.Statement:max volume of rect. prismFormula: xxyPersonalized Formula:18Example 3:Find the dimensions of the largest rectangle that can be inscribed in the ellipse in such a way that the sides are parallel to the axes .

ellipseMajor axis y-axis length is 2Minor axis x-axis length is 1Max Area: A=lwA=2x*2yA=4xySecondary equations19Example 4:Find the point on closest to the point (0, -1).

min(0,-1)20Example 5:A closed box with a square base is to have a volume of 2000in.3 . the material on the top and bottom will cost 3 cents per square inch and the material on the sides will cost 1 cent per square inch. Find the dimensions that will minimize the cost. Statement:minimize cost of materialFormula:xxy21Example 6:Suppose that P(x), R(x), and C(x) are the profit, revenue, and cost functions, that P(x) = R(x) - C(x), and x represents thousand of units.

Find the production level that maximizes the profit.

22Example 7:AP Type Problem:At noon a sailboat is 20 km south of a freighter. The sailboat is traveling east at 20 km/hr, and the freighter is traveling south at 40 km/hr. If the visibility is 10 km, could the people on the ships see each other? 23Example 8:AP - Max/min - Related RatesThe cross section of a trough has the shape of an inverted isosceles triangle. The lengths of the sides of the cross section are 15 in., and the length of the trough is 120 in.15in.120 in.1) Find the size of the vertex angle that will give the maximum capacity of the trough.2) If water is being added to the trough at 36 in3/min, how fast is the water level rising when the level is 5 in high? 24Last Update:12/03/10Assignment: DWK 4.425