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MAT 211- Lecture 6

by: Brigette Maggio

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MAT 211- Lecture 6 MAT 211

Marketplace > Arizona State University > Math > MAT 211 > MAT 211 Lecture 6
Brigette Maggio
ASU
GPA 3.9

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These notes cover MAT 211- Lecture 6 on Sections 15.3 and 15.4.
COURSE
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Dr. Mohacsy
TYPE
Class Notes
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2
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Math, MAT 211
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This 2 page Class Notes was uploaded by Brigette Maggio on Friday January 29, 2016. The Class Notes belongs to MAT 211 at Arizona State University taught by Dr. Mohacsy in Fall 2015. Since its upload, it has received 29 views. For similar materials see Math for Business Analysis in Math at Arizona State University.

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Date Created: 01/29/16
MAT 211­ Lecture 6­ 15.3 & 15.4  Section 15.3    Homework #8: ln(x,y)= x​ +y​2­y​x­2  lnx= 0 & lny= 0    1st Derivative:​ lnx= 2x­y​  lny= 2y­2yx  I. 2x­y​=0 ­­­­­­­> x=1 2­y​=0 2=y​ y=­ 2  2​ 2y­2yx=0 ­­­­­­­> y=0 2x­y​ =0 2x­0=0 x=0    II. 2y(1­x)= 0 y=0 & x=1    Critical Points: (0,0); (1, ­ 2); (1, + 2)    2nd Derivative:​  lnxx= 2 lnxx (0,0)= 2  lnyy= 2­2x lnyy (0,0)= 2  lnxy= ­2y lnxy (0,0)= 0    H(0,0)= 2 * 2 ­ 0= 4 Since lnxx= 2>0, then we have a relative minimum at (0,0).    ­Critical point at (1, ­ 2) : lnxx (1, ­ 2) =2   √ lnyy (1, ­ 2) =0  lnxy (1, ­ 2) =(­2)(­ 2) = 2 2  2  H (1, ­ 2)= 2 * 0 ­ (2 2)​          0­8= ­8< 0 Since H is negative, we have a saddle point at (1, ­ 2).    ­Critical point at (1, + 2) : lnxx (1, + 2)=2    lnyy (1, + 2)=0  lnxy (1, + 2)= ­2( 2)  2​ H(1, + 2) = 2* 0­ (­2 2)​=  0­8= ­8<0 Since H is negative, we have a saddle point at  (1, + 2).    Homework #11: M(c,f)= 11+8c+4c​ 2­40f­20cf+40f​2  Mc= 8+8c­20f­­­­­­­> 8+8c­20f= 0 divide by 4  2+2c­5f=0  Mf= ­40­20c+80f­­­­> ­40­20c+80f=0 divide by 20  ­2­c+4f=0    2+2c­5f=0 ­­­­­>2+2c­5f=0+       (2) ­2­c+4f=0 ­­­­­­>­4­2c+8f=0   ­2+3f=0  3f=2  f= ⅔  Plug back into:  ​2+2c­5(⅔)= 0     2+2c­10/3=0     2c= ­2+ 10/3 = 4/3     c= 4/3 * ½= ⅔=c  Critical point at (⅔, ⅔)    15.4 (#1,2,&3)­ Constrained Maxima and Minima  Find the minimum value of f(x,y,)= 2x​2+2x+y​2­y+z​­7  Subject to: z=2y (Constraint)    Substitution Method:  Substitute z= 2y into f(x,y,z)    f(x,y)= 2x+2x+y​2­y­(2y)​­2y­7  2                     +4y​ 2x​2+2x+5y​2­3y­7 NO z here!    Find Critical points and classify them.    f(x,y)= 2x+2x+5y​ 2­3y­7  fx= 4x+2 ­­­­­­­> 4x+2=0 ­­­­­­­> x=­1/2  fy= 10y­3 ­­­­­­­> 10y­3=0 ­­­­­­> y= 3/10  z=2y= 2 (3/10)= 6/10= 3/5    fxx= 4 fxx (­½, 3/10)= 4  fyy= 10 fyy (­½, 3/10)= 10  fxy= 0 fxy (­½, 3/10)= 0  H(­½, 3/10)= 4*10­0​2= 40 > 0  If positive and fxx=4 > 0, we have a relative minimum at (­½, 3/10, ⅗ ).

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