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anonymous
 one year ago
Help would be appreciated! Will Medal!
State the minimum and maximum values (if any) of the linear expression under the given constraints:
1. 0<=y<=6x
0<=x<=4
x+3y
anonymous
 one year ago
Help would be appreciated! Will Medal! State the minimum and maximum values (if any) of the linear expression under the given constraints: 1. 0<=y<=6x 0<=x<=4 x+3y

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If it helps this is Algebra.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0An analysis of the inequalities by the Mathematica program is attached.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I still dont understand what to do with that information

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It just rewrites the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand what the third line is saying

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did the 4+3*2=10 come from and what does that mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would x=4 and y=2 be the maximum while x=0 and y=6 be the minimum or vice versa? And when the equation is graphed does it curve?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Please I still need help.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why have you been typing for so long but nothing pops up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x + 3y ( 0<=y<=2 && 0<=x<=4)  (2<y<6 && 0<=x<=6y)  (y==6 && x==0 ) When y is 6 then x has to be 0 from (y==6 && x==0 ) . Then x+3y is 0 + 3*6 = 18, a maximum value. From ( 0 <= y <= 2 && 0 <= x <= 4 ) , x and y can be zero at the same time making x + 3y = 0, a minimum.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is about all I can do with this problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how do you graph it though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And thanks for your help so far

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't think I can help you on the graphing at this time. Sorry about that.
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