H huthaifa1984 Oct 2019 7 0 iraq Oct 30, 2019 #1 solve the following question and your will enter the heaven Attachments H. W -2 Optimization491952057470656799.pdf.doc 18.7 KB Views: 9

skipjack Forum Staff Dec 2006 21,379 2,410 Oct 30, 2019 #2 Did you copy and paste that into Word from a pdf file? Unfortunately, that makes it pretty much unreadable on my PC. Reactions: 2 people

Did you copy and paste that into Word from a pdf file? Unfortunately, that makes it pretty much unreadable on my PC.

H huthaifa1984 Oct 2019 7 0 iraq Oct 31, 2019 #3 zip here zip file Attachments Untitled11.zip 88.8 KB Views: 4

tahirimanov19 Mar 2015 182 68 Universe 2.71828i3.14159 Oct 31, 2019 #4 These are easy problems. Confused about Q1, integrate between -1 and y?

skeeter Math Team Jul 2011 3,266 1,766 Texas Oct 31, 2019 #5 As a courtesy to the OP ... Reactions: 2 people

D DarnItJimImAnEngineer Jun 2019 493 262 USA Oct 31, 2019 #6 I'm confused about Q1. Where are x and y supposed to be the specified function arguments and where are they supposed to be the variables of integration?

I'm confused about Q1. Where are x and y supposed to be the specified function arguments and where are they supposed to be the variables of integration?

tahirimanov19 Mar 2015 182 68 Universe 2.71828i3.14159 Oct 31, 2019 #7 Q2. $(x+y+z) = \dfrac{x+y}{2} + \dfrac{y+z}{2} + \dfrac{z+x}{2} \ge \sqrt{xy} + \sqrt{yz} + \sqrt{zx} \ge \sqrt{xy+yz+zx}$ Q3. y and z in terms of x, and for $x \in Z^+$ find the minimum value of f(x). Q4 is too easy. And there is still something wrong with Q1. Last edited by a moderator: Nov 1, 2019

Q2. $(x+y+z) = \dfrac{x+y}{2} + \dfrac{y+z}{2} + \dfrac{z+x}{2} \ge \sqrt{xy} + \sqrt{yz} + \sqrt{zx} \ge \sqrt{xy+yz+zx}$ Q3. y and z in terms of x, and for $x \in Z^+$ find the minimum value of f(x). Q4 is too easy. And there is still something wrong with Q1.

idontknow Dec 2015 972 128 Earth Nov 1, 2019 #9 Here : Calculus III - Lagrange Multipliers Reactions: 1 person