OPTIMIZATION PROBLEMS Maxim and minim interval closed [closed]
Find a number in the closed interval $[frac{1}{2},frac{3}{2}]$
such that the sum of the number and its reciprocal is
(a) as small as possible
(b) as large as possible.
optimization
edited Dec 28 '18 at 19:18
greelious
15510
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
closed as off-topic by José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo Dec 29 '18 at 1:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Find a number in the closed interval $[frac{1}{2},frac{3}{2}]$
such that the sum of the number and its reciprocal is
(a) as small as possible
(b) as large as possible.
optimization
edited Dec 28 '18 at 19:18
greelious
15510
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
closed as off-topic by José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo Dec 29 '18 at 1:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
What have you tried? Note: I assume you are referring to the open interval $(frac 12, frac 32)$, yes?
– lulu
Dec 28 '18 at 19:06
3
Done! What else do you want me to do?
– José Carlos Santos
Dec 28 '18 at 19:06
Hint: take a derivative.
– Sean Roberson
Dec 28 '18 at 19:26
add a comment |
Find a number in the closed interval $[frac{1}{2},frac{3}{2}]$
such that the sum of the number and its reciprocal is
(a) as small as possible
(b) as large as possible.
optimization
edited Dec 28 '18 at 19:18
greelious
15510
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
Find a number in the closed interval $[frac{1}{2},frac{3}{2}]$
such that the sum of the number and its reciprocal is
(a) as small as possible
(b) as large as possible.
optimization
optimization
edited Dec 28 '18 at 19:18
greelious
15510
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
edited Dec 28 '18 at 19:18
greelious
15510
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
edited Dec 28 '18 at 19:18
greelious
15510
edited Dec 28 '18 at 19:18
greelious
15510
edited Dec 28 '18 at 19:18
greelious
15510
15510
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
asked Dec 28 '18 at 19:02
Jaime PérezJaime Pérez
114
114
closed as off-topic by José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo Dec 29 '18 at 1:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo Dec 29 '18 at 1:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Cesareo, Eevee Trainer, Leucippus, Holo
If this question can be reworded to fit the rules in the help center, please edit the question.
What have you tried? Note: I assume you are referring to the open interval $(frac 12, frac 32)$, yes?
– lulu
Dec 28 '18 at 19:06
3
Done! What else do you want me to do?
– José Carlos Santos
Dec 28 '18 at 19:06
Hint: take a derivative.
– Sean Roberson
Dec 28 '18 at 19:26
add a comment |
What have you tried? Note: I assume you are referring to the open interval $(frac 12, frac 32)$, yes?
– lulu
Dec 28 '18 at 19:06
3
Done! What else do you want me to do?
– José Carlos Santos
Dec 28 '18 at 19:06
Hint: take a derivative.
– Sean Roberson
Dec 28 '18 at 19:26
What have you tried? Note: I assume you are referring to the open interval $(frac 12, frac 32)$, yes?
– lulu
Dec 28 '18 at 19:06
What have you tried? Note: I assume you are referring to the open interval $(frac 12, frac 32)$, yes?
– lulu
Dec 28 '18 at 19:06
3
3
Done! What else do you want me to do?
– José Carlos Santos
Dec 28 '18 at 19:06
Done! What else do you want me to do?
– José Carlos Santos
Dec 28 '18 at 19:06
Hint: take a derivative.
– Sean Roberson
Dec 28 '18 at 19:26
Hint: take a derivative.
– Sean Roberson
Dec 28 '18 at 19:26
add a comment |
3 Answers
3
active
oldest
votes
when $x=0.5$,then $z_{max}=2.5$
and when $x=1$,then $z_{min}=2$
I did it by linear programming.just plot the function $z=x+frac{1}{x}$ and the area of $frac{1}{2}le xlefrac{3}{2}$
and check the intersected corner point's $~~"x"~~$ value.and plug it into the function to get the corresponding value of $z$ .
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
add a comment |
$frac{d}{dx}left(x+frac{1}{x}right)=1-frac{1}{x^2}$
$$0=1-frac{1}{x^2}$$
$$ x=1~~or,~~ x=-1 $$
but as the interval is $0.5≤x≤1.5$
we only take $x = 1$
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
add a comment |
The second derivative of $f(x):=x+x^{-1}$, which is $f''(x)=2/x^{3}$, is strictly positive on $(0.5, 1.5)$. This means $f$ is strictly convex on this interval. The critical point $x=1$, found by solving $f'(x)=0$, is therefore a global minimum of $f$ on $[0.5,1.5]$.
To find the maximum, compare $f(0.5)$ against $f(1.5)$. Clearly, $f(0.5)=0.5+2=2.5>1.5+2/3=f(1.5)$.
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
when $x=0.5$,then $z_{max}=2.5$
and when $x=1$,then $z_{min}=2$
I did it by linear programming.just plot the function $z=x+frac{1}{x}$ and the area of $frac{1}{2}le xlefrac{3}{2}$
and check the intersected corner point's $~~"x"~~$ value.and plug it into the function to get the corresponding value of $z$ .
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
add a comment |
when $x=0.5$,then $z_{max}=2.5$
and when $x=1$,then $z_{min}=2$
I did it by linear programming.just plot the function $z=x+frac{1}{x}$ and the area of $frac{1}{2}le xlefrac{3}{2}$
and check the intersected corner point's $~~"x"~~$ value.and plug it into the function to get the corresponding value of $z$ .
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
add a comment |
when $x=0.5$,then $z_{max}=2.5$
and when $x=1$,then $z_{min}=2$
I did it by linear programming.just plot the function $z=x+frac{1}{x}$ and the area of $frac{1}{2}le xlefrac{3}{2}$
and check the intersected corner point's $~~"x"~~$ value.and plug it into the function to get the corresponding value of $z$ .
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
when $x=0.5$,then $z_{max}=2.5$
and when $x=1$,then $z_{min}=2$
I did it by linear programming.just plot the function $z=x+frac{1}{x}$ and the area of $frac{1}{2}le xlefrac{3}{2}$
and check the intersected corner point's $~~"x"~~$ value.and plug it into the function to get the corresponding value of $z$ .
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
edited Dec 28 '18 at 19:23
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:13
Rakibul Islam PrinceRakibul Islam Prince
1,035211
1,035211
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
add a comment |
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
that is not linear programming
– LinAlg
Dec 28 '18 at 20:16
add a comment |
$frac{d}{dx}left(x+frac{1}{x}right)=1-frac{1}{x^2}$
$$0=1-frac{1}{x^2}$$
$$ x=1~~or,~~ x=-1 $$
but as the interval is $0.5≤x≤1.5$
we only take $x = 1$
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
add a comment |
$frac{d}{dx}left(x+frac{1}{x}right)=1-frac{1}{x^2}$
$$0=1-frac{1}{x^2}$$
$$ x=1~~or,~~ x=-1 $$
but as the interval is $0.5≤x≤1.5$
we only take $x = 1$
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
add a comment |
$frac{d}{dx}left(x+frac{1}{x}right)=1-frac{1}{x^2}$
$$0=1-frac{1}{x^2}$$
$$ x=1~~or,~~ x=-1 $$
but as the interval is $0.5≤x≤1.5$
we only take $x = 1$
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
$frac{d}{dx}left(x+frac{1}{x}right)=1-frac{1}{x^2}$
$$0=1-frac{1}{x^2}$$
$$ x=1~~or,~~ x=-1 $$
but as the interval is $0.5≤x≤1.5$
we only take $x = 1$
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
edited Dec 28 '18 at 20:08
Rakibul Islam Prince
1,035211
1,035211
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
answered Dec 28 '18 at 19:45
Jaime PérezJaime Pérez
114
114
add a comment |
add a comment |
The second derivative of $f(x):=x+x^{-1}$, which is $f''(x)=2/x^{3}$, is strictly positive on $(0.5, 1.5)$. This means $f$ is strictly convex on this interval. The critical point $x=1$, found by solving $f'(x)=0$, is therefore a global minimum of $f$ on $[0.5,1.5]$.
To find the maximum, compare $f(0.5)$ against $f(1.5)$. Clearly, $f(0.5)=0.5+2=2.5>1.5+2/3=f(1.5)$.
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
add a comment |
The second derivative of $f(x):=x+x^{-1}$, which is $f''(x)=2/x^{3}$, is strictly positive on $(0.5, 1.5)$. This means $f$ is strictly convex on this interval. The critical point $x=1$, found by solving $f'(x)=0$, is therefore a global minimum of $f$ on $[0.5,1.5]$.
To find the maximum, compare $f(0.5)$ against $f(1.5)$. Clearly, $f(0.5)=0.5+2=2.5>1.5+2/3=f(1.5)$.
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
add a comment |
The second derivative of $f(x):=x+x^{-1}$, which is $f''(x)=2/x^{3}$, is strictly positive on $(0.5, 1.5)$. This means $f$ is strictly convex on this interval. The critical point $x=1$, found by solving $f'(x)=0$, is therefore a global minimum of $f$ on $[0.5,1.5]$.
To find the maximum, compare $f(0.5)$ against $f(1.5)$. Clearly, $f(0.5)=0.5+2=2.5>1.5+2/3=f(1.5)$.
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
The second derivative of $f(x):=x+x^{-1}$, which is $f''(x)=2/x^{3}$, is strictly positive on $(0.5, 1.5)$. This means $f$ is strictly convex on this interval. The critical point $x=1$, found by solving $f'(x)=0$, is therefore a global minimum of $f$ on $[0.5,1.5]$.
To find the maximum, compare $f(0.5)$ against $f(1.5)$. Clearly, $f(0.5)=0.5+2=2.5>1.5+2/3=f(1.5)$.
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
edited Dec 31 '18 at 1:35
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
answered Dec 29 '18 at 1:19
J.DoeJ.Doe
63
63
add a comment |
add a comment |
What have you tried? Note: I assume you are referring to the open interval $(frac 12, frac 32)$, yes?
– lulu
Dec 28 '18 at 19:06
3
Done! What else do you want me to do?
– José Carlos Santos
Dec 28 '18 at 19:06
Hint: take a derivative.
– Sean Roberson
Dec 28 '18 at 19:26