Minimum value of a linear program












2












$begingroup$


$x$ and $y$ are real numbers such that $ xge 0$ and $y ge 0$.



If $$ x + y le 5$$
$$ x + 2y ge 8$$



Then what is the minimum value of $5x+y$?



There is something wrong in my approach.
I write the first inequality as $$ 5 ge x + y$$
Adding with the second inequality yields, $$5 + x + 2y ge x + y + 8$$
Or $$ y ge 3$$ Thus $$ 9y ge 27$$
Multiplying the second inequality by $5$ we obtain $$ 5x + 10y ge 40$$ $$9y ge 27$$
Hence $5x+y ge 13$






inequality linear-programming







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You can add the equations $5x+10y geq 40$ and $9y geq 27$ to conclude $5 x+19 y geq 67$ but you can't subtract them and conclude anything meaningful. For example, we have $5 geq 3$ and $5 geq 2$ but you can't subtract the second from the first and say $0 geq 1$.
    $endgroup$
    – Eric
    Jan 10 at 11:27










  • $begingroup$
    So, how do you find the minimum value?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:30










  • $begingroup$
    @Atiq Rahman I think, you are right. $13$ is a minimal value.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:35






  • 1




    $begingroup$
    I got $$5x+ygeq 4$$ for $x=0,y=4$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 10 at 11:43










  • $begingroup$
    $4$ is the minimum value @MichaelRozenberg . $(x,y)=(0,4)$ satisfies those inequalities.
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:43
















2












$begingroup$


$x$ and $y$ are real numbers such that $ xge 0$ and $y ge 0$.



If $$ x + y le 5$$
$$ x + 2y ge 8$$



Then what is the minimum value of $5x+y$?



There is something wrong in my approach.
I write the first inequality as $$ 5 ge x + y$$
Adding with the second inequality yields, $$5 + x + 2y ge x + y + 8$$
Or $$ y ge 3$$ Thus $$ 9y ge 27$$
Multiplying the second inequality by $5$ we obtain $$ 5x + 10y ge 40$$ $$9y ge 27$$
Hence $5x+y ge 13$






inequality linear-programming







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You can add the equations $5x+10y geq 40$ and $9y geq 27$ to conclude $5 x+19 y geq 67$ but you can't subtract them and conclude anything meaningful. For example, we have $5 geq 3$ and $5 geq 2$ but you can't subtract the second from the first and say $0 geq 1$.
    $endgroup$
    – Eric
    Jan 10 at 11:27










  • $begingroup$
    So, how do you find the minimum value?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:30










  • $begingroup$
    @Atiq Rahman I think, you are right. $13$ is a minimal value.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:35






  • 1




    $begingroup$
    I got $$5x+ygeq 4$$ for $x=0,y=4$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 10 at 11:43










  • $begingroup$
    $4$ is the minimum value @MichaelRozenberg . $(x,y)=(0,4)$ satisfies those inequalities.
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:43














2












2








2





$begingroup$


$x$ and $y$ are real numbers such that $ xge 0$ and $y ge 0$.



If $$ x + y le 5$$
$$ x + 2y ge 8$$



Then what is the minimum value of $5x+y$?



There is something wrong in my approach.
I write the first inequality as $$ 5 ge x + y$$
Adding with the second inequality yields, $$5 + x + 2y ge x + y + 8$$
Or $$ y ge 3$$ Thus $$ 9y ge 27$$
Multiplying the second inequality by $5$ we obtain $$ 5x + 10y ge 40$$ $$9y ge 27$$
Hence $5x+y ge 13$






inequality linear-programming







share|cite|improve this question









$endgroup$




$x$ and $y$ are real numbers such that $ xge 0$ and $y ge 0$.



If $$ x + y le 5$$
$$ x + 2y ge 8$$



Then what is the minimum value of $5x+y$?



There is something wrong in my approach.
I write the first inequality as $$ 5 ge x + y$$
Adding with the second inequality yields, $$5 + x + 2y ge x + y + 8$$
Or $$ y ge 3$$ Thus $$ 9y ge 27$$
Multiplying the second inequality by $5$ we obtain $$ 5x + 10y ge 40$$ $$9y ge 27$$
Hence $5x+y ge 13$






inequality linear-programming




inequality linear-programming






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 11:19









Atiq RahmanAtiq Rahman

1063




1063








  • 1




    $begingroup$
    You can add the equations $5x+10y geq 40$ and $9y geq 27$ to conclude $5 x+19 y geq 67$ but you can't subtract them and conclude anything meaningful. For example, we have $5 geq 3$ and $5 geq 2$ but you can't subtract the second from the first and say $0 geq 1$.
    $endgroup$
    – Eric
    Jan 10 at 11:27










  • $begingroup$
    So, how do you find the minimum value?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:30










  • $begingroup$
    @Atiq Rahman I think, you are right. $13$ is a minimal value.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:35






  • 1




    $begingroup$
    I got $$5x+ygeq 4$$ for $x=0,y=4$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 10 at 11:43










  • $begingroup$
    $4$ is the minimum value @MichaelRozenberg . $(x,y)=(0,4)$ satisfies those inequalities.
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:43














  • 1




    $begingroup$
    You can add the equations $5x+10y geq 40$ and $9y geq 27$ to conclude $5 x+19 y geq 67$ but you can't subtract them and conclude anything meaningful. For example, we have $5 geq 3$ and $5 geq 2$ but you can't subtract the second from the first and say $0 geq 1$.
    $endgroup$
    – Eric
    Jan 10 at 11:27










  • $begingroup$
    So, how do you find the minimum value?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:30










  • $begingroup$
    @Atiq Rahman I think, you are right. $13$ is a minimal value.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:35






  • 1




    $begingroup$
    I got $$5x+ygeq 4$$ for $x=0,y=4$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 10 at 11:43










  • $begingroup$
    $4$ is the minimum value @MichaelRozenberg . $(x,y)=(0,4)$ satisfies those inequalities.
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:43








1




1




$begingroup$
You can add the equations $5x+10y geq 40$ and $9y geq 27$ to conclude $5 x+19 y geq 67$ but you can't subtract them and conclude anything meaningful. For example, we have $5 geq 3$ and $5 geq 2$ but you can't subtract the second from the first and say $0 geq 1$.
$endgroup$
– Eric
Jan 10 at 11:27




$begingroup$
You can add the equations $5x+10y geq 40$ and $9y geq 27$ to conclude $5 x+19 y geq 67$ but you can't subtract them and conclude anything meaningful. For example, we have $5 geq 3$ and $5 geq 2$ but you can't subtract the second from the first and say $0 geq 1$.
$endgroup$
– Eric
Jan 10 at 11:27












$begingroup$
So, how do you find the minimum value?
$endgroup$
– Atiq Rahman
Jan 10 at 11:30




$begingroup$
So, how do you find the minimum value?
$endgroup$
– Atiq Rahman
Jan 10 at 11:30












$begingroup$
@Atiq Rahman I think, you are right. $13$ is a minimal value.
$endgroup$
– Michael Rozenberg
Jan 10 at 11:35




$begingroup$
@Atiq Rahman I think, you are right. $13$ is a minimal value.
$endgroup$
– Michael Rozenberg
Jan 10 at 11:35




1




1




$begingroup$
I got $$5x+ygeq 4$$ for $x=0,y=4$
$endgroup$
– Dr. Sonnhard Graubner
Jan 10 at 11:43




$begingroup$
I got $$5x+ygeq 4$$ for $x=0,y=4$
$endgroup$
– Dr. Sonnhard Graubner
Jan 10 at 11:43












$begingroup$
$4$ is the minimum value @MichaelRozenberg . $(x,y)=(0,4)$ satisfies those inequalities.
$endgroup$
– Atiq Rahman
Jan 10 at 11:43




$begingroup$
$4$ is the minimum value @MichaelRozenberg . $(x,y)=(0,4)$ satisfies those inequalities.
$endgroup$
– Atiq Rahman
Jan 10 at 11:43










1 Answer
1






active

oldest

votes


















1












$begingroup$

Our conditions give an interior of $Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$



The system $x=0$ and $x+2y=8$ gives $C(0,4)$;



The system $x=0$ and $x+y=5$ gives $B(0,5)$



and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$



Let $f(x,y)=5x+y.$



Thus, $$min_{xgeq0,ygeq0,x+yleq5,x+2ygeq8}f=min{f(0,5),f(0,4),f(2,3)}=f(0,4)=4.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you find the points $(0,5),(0,4),(2,3)$?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:53










  • $begingroup$
    @Atiq Rahman I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:57












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Our conditions give an interior of $Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$



The system $x=0$ and $x+2y=8$ gives $C(0,4)$;



The system $x=0$ and $x+y=5$ gives $B(0,5)$



and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$



Let $f(x,y)=5x+y.$



Thus, $$min_{xgeq0,ygeq0,x+yleq5,x+2ygeq8}f=min{f(0,5),f(0,4),f(2,3)}=f(0,4)=4.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you find the points $(0,5),(0,4),(2,3)$?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:53










  • $begingroup$
    @Atiq Rahman I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:57
















1












$begingroup$

Our conditions give an interior of $Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$



The system $x=0$ and $x+2y=8$ gives $C(0,4)$;



The system $x=0$ and $x+y=5$ gives $B(0,5)$



and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$



Let $f(x,y)=5x+y.$



Thus, $$min_{xgeq0,ygeq0,x+yleq5,x+2ygeq8}f=min{f(0,5),f(0,4),f(2,3)}=f(0,4)=4.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you find the points $(0,5),(0,4),(2,3)$?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:53










  • $begingroup$
    @Atiq Rahman I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:57














1












1








1





$begingroup$

Our conditions give an interior of $Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$



The system $x=0$ and $x+2y=8$ gives $C(0,4)$;



The system $x=0$ and $x+y=5$ gives $B(0,5)$



and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$



Let $f(x,y)=5x+y.$



Thus, $$min_{xgeq0,ygeq0,x+yleq5,x+2ygeq8}f=min{f(0,5),f(0,4),f(2,3)}=f(0,4)=4.$$






share|cite|improve this answer











$endgroup$



Our conditions give an interior of $Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$



The system $x=0$ and $x+2y=8$ gives $C(0,4)$;



The system $x=0$ and $x+y=5$ gives $B(0,5)$



and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$



Let $f(x,y)=5x+y.$



Thus, $$min_{xgeq0,ygeq0,x+yleq5,x+2ygeq8}f=min{f(0,5),f(0,4),f(2,3)}=f(0,4)=4.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 11:56

























answered Jan 10 at 11:49









Michael RozenbergMichael Rozenberg

110k1896201




110k1896201












  • $begingroup$
    How do you find the points $(0,5),(0,4),(2,3)$?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:53










  • $begingroup$
    @Atiq Rahman I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:57


















  • $begingroup$
    How do you find the points $(0,5),(0,4),(2,3)$?
    $endgroup$
    – Atiq Rahman
    Jan 10 at 11:53










  • $begingroup$
    @Atiq Rahman I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 10 at 11:57
















$begingroup$
How do you find the points $(0,5),(0,4),(2,3)$?
$endgroup$
– Atiq Rahman
Jan 10 at 11:53




$begingroup$
How do you find the points $(0,5),(0,4),(2,3)$?
$endgroup$
– Atiq Rahman
Jan 10 at 11:53












$begingroup$
@Atiq Rahman I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 10 at 11:57




$begingroup$
@Atiq Rahman I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 10 at 11:57


















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