Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

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<=6-x 0<=x<=4 x+3y

OpenStudy (anonymous):

@robtobey

OpenStudy (anonymous):

Any ideas?

OpenStudy (anonymous):

If it helps this is Algebra.

OpenStudy (anonymous):

An analysis of the inequalities by the Mathematica program is attached.

OpenStudy (anonymous):

I still dont understand what to do with that information

OpenStudy (anonymous):

It just rewrites the question

OpenStudy (anonymous):

I don't understand what the third line is saying

OpenStudy (anonymous):

where did the 4+3*2=10 come from and what does that mean?

OpenStudy (anonymous):

Would 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?

OpenStudy (anonymous):

Please I still need help.

OpenStudy (anonymous):

why have you been typing for so long but nothing pops up.

OpenStudy (anonymous):

x + 3y ( 0<=y<=2 && 0<=x<=4) || (2<y<6 && 0<=x<=6-y) || (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.

OpenStudy (anonymous):

That is about all I can do with this problem.

OpenStudy (anonymous):

So how do you graph it though

OpenStudy (anonymous):

And thanks for your help so far

OpenStudy (anonymous):

Don't think I can help you on the graphing at this time. Sorry about that.

OpenStudy (anonymous):

thanks anyway

OpenStudy (anonymous):

Have good day.

OpenStudy (anonymous):

Bye

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!