Review definite integral and Riemann Sum before 12.1, read page 829

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1 Review definite integral Riemann Sum before 1.1, read page LaGrange Multiples Lagrange s method is a wa to find maximum minimum values of a general function subject to certain constraints (conditions). It can sometimes be used as an alternative to the Second Derivatives Test from Section 11.7 sometimes in the case where the value of D= Let s start with a function of two variables f = f( x, ). Method of Lagrange Multipliers To find the maximum minimum values of f ( x, ) subject to constraint g ( x, ) a. find all values of x,, λ so that f ( x, ) = λ g( x, ) g( x, ) (This will be our sstem of equations to solve) b. Evaluate f at all the points ( x, ) ou found in step a). The largest will obviousl be the maximum value of f the smallest will be the minimum value of f TEC 11.8 Blue curve is constraint g( x, ) (k is a constant so make sure ou have a constant there) Red plane different level curves of the function Gre is the 3D surface f = f( x, ) Level curves is in Part A Level surfaces (which we ll do later toda) is in part B. Part A the surface f ( x, ) will have level curves which are shown in red. This shows the function(surface) has maximum/minimum values when the red level curves are tangent to the blue constraints curve, when the SHARE a common tangent line. The are a multiple of each other we use lambda, λ, as the multiple. g(x,) f f is parallel to g - the gradient vectors are parallel to each other at the point of tangenc Here in red are 3 level curves of f (x,). f = λ g g Scalar multiple (just a #) common tangent lambda Lagrange Multiplier line

2 I HW problem #4, we are asked to use Lagrange multipliers to find the maximum minimum values f x, = 4x+ 6subject to constraint x + = 13. of TEC 11.8 Part A pretend the red plane in TEC is f x, = 4x+ 6 we re tring to see the maximum/minimum of f with constraint in blue. Or put it on Mathematica. Before, we would need to solve x 13 f x, = 4x+ 6to get one function f. Back in section 11.7 to find max./min. values, we had one function f = f( x, ) 1. found critical points where f = f = 0, then. found value of D f f ( f ) + = for either variable substitute it in to x xx x = using the second derivatives test determined of Local minimum ( > 0, > 0), Local maximum ( > 0, < 0) D f xx D f xx I tried this one with the 11.7 method but it failed to be useful because D=0. f( x, ) = 4 x + 6 f x f = 4 ± f =± f =± = 0 f =± Before we find the points let's look at the second derivatives test: Dx (, ) = f f f xx x Dx (, ) = 0 f 0 = 0 So the test fails. Now we re using Lagrange multipliers to find maximum/minimum values of f = f( x, ) subject to a g. You will be redoing man problems from 11.7 on 11.8 using Lagrange multipliers. constraint Instead of having one function f f( x, ) HW #4. =, ou ll have it along with a constraint Use Lagrange multipliers to find the maximum minimum values of constraint x + = 13. g. f x, = 4x+ 6subject to

3 *Show the visual of #4 on Mathematica on our website Let s move up a level Same idea, but increase function f ( x, ) of two variables to f (,, ) Method of Lagrange Multipliers xz, a function of three variables: To find the maximum minimum values of f ( xz,, ) subject to constraint g ( xz,, ) a. find all values of x,, z, λ so that f ( x,, z) = λ g( x,, z) g( x,, z) (This will be our sstem of equations to solve) b. Evaluate f at all the points ( x, z, ) ou found in step a). The largest will obviousl be the maximum value of f the smallest will be the minimum value of f TEC 11.8 Part B The 4 th dimensional function f (,, ) The Blue is the constraint surface gxz (,, ) xz will have level surfaces which are shown in red.. This shows the maximums/minimums of f (,, ) xz occur where the red level surfaces are tangent to the blue constraint surface. The share a common tangent plane. Maneuver it around to show where. Again, gradient f gradient g will be parallel to each other, a scalar multiple of each other: f = λ g Remember 11.7 notes Example 5 where we found the points on the surface the origin. We substituted = 9 + xz into = 9 + xz that were closest to d = x + + z found the minimum of our new function d( x, z) = x + 9+ xz+ z so we combined them into one function found the minimum of d( x, z. ) We found ( 0, ± 3,0) were the points that minimized that distance I showed the visual on Mathematica. Did I convince ou to just minimize d( x, z) = x + 9+ xz+ z instead of d( x, z) = x + 9+ xz+ z? Let s redo this problem from 11.7, which is also a problem on Now we keep = 9 + xz d = x + + z separate. We want to minimize d( x, z ) subject to constraint curve xz = 9 or 9 xz = 0 gxz (,, ), so either leave the 9 on the right side or bring everthing over have a 0 on the right side.

4 Example: Use Lagrange multipliers to find the points on the surface graphing calculator to solve the sstem of equations. = 9 + xz that are closest to the origin. Use a Find minimum d = x + + z or d = f x,, z = x + + z g x,, z = xz = 9 with constraint g(x,,z)=k You ll notice the different set up in 11.8 than in Back in 11.7 we took ^ x z = 9 solved it for ^ plugged it back into d in order to get one fn. Now we leave them separate.

5 nd CHAR Greek; A On TI-89 nd MATH, 8, 8 Solve ( x= λ z = λ z = λ x x z = 9, { x,, z, λ} ) Make sure ou use the multiplication smbol between variables. (or ou can leave λ off) λ s kind of a pain to get, ou could just use T, but I like to see it. Be able to solve these HW problems manuall AND with our graphing calculator; the ma be on either part of the test. I distinguished the tpes on HW I would expect to solve manuall. Let s tr one last example solve it manuall: HW #8 Use Lagrange multipliers to find the maximum minimum values of the function subject to the given constraint x z = 5. f xz,, = 8x 4z So, given an optimization problem ou can either use: 1) 11.7 s method of finding critical points, finding D, then testing the critical points at D (ou need one function f), or ) 11.8 s method of Lagrange Multipliers

FINAL EXAM STUDY GUIDE

FINAL EXAM STUDY GUIDE The Final Exam takes place on Wednesday, June 13, 2018, from 10:30 AM to 12:30 PM in 1100 Donald Bren Hall (not the usual lecture room!!!) NO books/notes/calculators/cheat sheets