How to correctly find the value of theta for which $fracacostheta+fracbsintheta$ is minimum? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that minimum of $lambda sin theta + (1 - lambda) cos theta le -frac1sqrt 2$How can I prove the trigonometric Problem?If $x,y in (0,fracpi2)$ then expression $sin x +cos y +tan^2y+cot^2x+5>ldots?$Conditional inequalityFind the maximum and minimum values of $sin^2theta+sin^2phi$ when $theta+phi=alpha$Prove that $-4leq5costheta+3cos(theta+fracpi3)+3leq10$Problem in finding the maximum value of $tan^2(theta-phi)$Find the minimum of the value $f=(1+sin^2x)(1+sin^2y)$Two different answers by applying AM GM InequalityFind the minimum value of $sin^2theta+cos^2theta+csc^2theta+sec^2theta+tan^2theta+cot^2 theta$
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How to correctly find the value of theta for which $fracacostheta+fracbsintheta$ is minimum?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that minimum of $lambda sin theta + (1 - lambda) cos theta le -frac1sqrt 2$How can I prove the trigonometric Problem?If $x,y in (0,fracpi2)$ then expression $sin x +cos y +tan^2y+cot^2x+5>ldots?$Conditional inequalityFind the maximum and minimum values of $sin^2theta+sin^2phi$ when $theta+phi=alpha$Prove that $-4leq5costheta+3cos(theta+fracpi3)+3leq10$Problem in finding the maximum value of $tan^2(theta-phi)$Find the minimum of the value $f=(1+sin^2x)(1+sin^2y)$Two different answers by applying AM GM InequalityFind the minimum value of $sin^2theta+cos^2theta+csc^2theta+sec^2theta+tan^2theta+cot^2 theta$
$begingroup$
I was solving a question which required me to find the value of $theta$ for which the expression $fracacostheta+fracbsintheta$ has its minimum value. Given that, $a=3sqrt3, b=1$.
Note: It was a fault from my end for not including the values of $a,b$ which were given in the question. I apologize for that.
Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following:
Using AM-GM inequality
$$fracacostheta+fracbsintheta ge 2sqrtfracabcosthetasintheta\ implies fracacostheta+fracbsintheta ge 2sqrtfrac2absin2theta$$
Since, we're minising the given expression, therefore $sin2theta$ should have maximum value, i.e. $sin2theta=1$. Therefore, $theta=dfracpi4$.
Using calculus
$$textLet f(theta)=asectheta+bcsctheta \
therefore f'(theta)=asecthetatantheta-bcscthetacottheta \
textFor mininma, f(theta)=0, implies asecthetatantheta=bcscthetacottheta \
textor, tan^3theta=fracba=frac13sqrt3 \
implies theta=fracpi6$$
Why does solving this problem using derivatives yield $theta=dfracpi6$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus?
inequality optimization a.m.-g.m.-inequality
edited Apr 8 at 14:36
Dr. Mathva
3,493630
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
$endgroup$
add a comment |
$begingroup$
I was solving a question which required me to find the value of $theta$ for which the expression $fracacostheta+fracbsintheta$ has its minimum value. Given that, $a=3sqrt3, b=1$.
Note: It was a fault from my end for not including the values of $a,b$ which were given in the question. I apologize for that.
Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following:
Using AM-GM inequality
$$fracacostheta+fracbsintheta ge 2sqrtfracabcosthetasintheta\ implies fracacostheta+fracbsintheta ge 2sqrtfrac2absin2theta$$
Since, we're minising the given expression, therefore $sin2theta$ should have maximum value, i.e. $sin2theta=1$. Therefore, $theta=dfracpi4$.
Using calculus
$$textLet f(theta)=asectheta+bcsctheta \
therefore f'(theta)=asecthetatantheta-bcscthetacottheta \
textFor mininma, f(theta)=0, implies asecthetatantheta=bcscthetacottheta \
textor, tan^3theta=fracba=frac13sqrt3 \
implies theta=fracpi6$$
Why does solving this problem using derivatives yield $theta=dfracpi6$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus?
inequality optimization a.m.-g.m.-inequality
edited Apr 8 at 14:36
Dr. Mathva
3,493630
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
$endgroup$
$begingroup$
Please show your work solving using derivatives; doesn't $theta$ depend on $a$ and $b$?
$endgroup$
– J. W. Tanner
Apr 8 at 14:11
$begingroup$
@J.W.Tanner Done!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:28
$begingroup$
Thanks for clarifying; you have the correct answer using calculus now
$endgroup$
– J. W. Tanner
Apr 8 at 14:37
$begingroup$
Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
$endgroup$
– Jon
Apr 8 at 14:54
add a comment |
$begingroup$
I was solving a question which required me to find the value of $theta$ for which the expression $fracacostheta+fracbsintheta$ has its minimum value. Given that, $a=3sqrt3, b=1$.
Note: It was a fault from my end for not including the values of $a,b$ which were given in the question. I apologize for that.
Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following:
Using AM-GM inequality
$$fracacostheta+fracbsintheta ge 2sqrtfracabcosthetasintheta\ implies fracacostheta+fracbsintheta ge 2sqrtfrac2absin2theta$$
Since, we're minising the given expression, therefore $sin2theta$ should have maximum value, i.e. $sin2theta=1$. Therefore, $theta=dfracpi4$.
Using calculus
$$textLet f(theta)=asectheta+bcsctheta \
therefore f'(theta)=asecthetatantheta-bcscthetacottheta \
textFor mininma, f(theta)=0, implies asecthetatantheta=bcscthetacottheta \
textor, tan^3theta=fracba=frac13sqrt3 \
implies theta=fracpi6$$
Why does solving this problem using derivatives yield $theta=dfracpi6$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus?
inequality optimization a.m.-g.m.-inequality
edited Apr 8 at 14:36
Dr. Mathva
3,493630
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
$endgroup$
I was solving a question which required me to find the value of $theta$ for which the expression $fracacostheta+fracbsintheta$ has its minimum value. Given that, $a=3sqrt3, b=1$.
Note: It was a fault from my end for not including the values of $a,b$ which were given in the question. I apologize for that.
Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following:
Using AM-GM inequality
$$fracacostheta+fracbsintheta ge 2sqrtfracabcosthetasintheta\ implies fracacostheta+fracbsintheta ge 2sqrtfrac2absin2theta$$
Since, we're minising the given expression, therefore $sin2theta$ should have maximum value, i.e. $sin2theta=1$. Therefore, $theta=dfracpi4$.
Using calculus
$$textLet f(theta)=asectheta+bcsctheta \
therefore f'(theta)=asecthetatantheta-bcscthetacottheta \
textFor mininma, f(theta)=0, implies asecthetatantheta=bcscthetacottheta \
textor, tan^3theta=fracba=frac13sqrt3 \
implies theta=fracpi6$$
Why does solving this problem using derivatives yield $theta=dfracpi6$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus?
inequality optimization a.m.-g.m.-inequality
inequality optimization a.m.-g.m.-inequality
edited Apr 8 at 14:36
Dr. Mathva
3,493630
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
edited Apr 8 at 14:36
Dr. Mathva
3,493630
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
edited Apr 8 at 14:36
Dr. Mathva
3,493630
edited Apr 8 at 14:36
Dr. Mathva
3,493630
edited Apr 8 at 14:36
Dr. Mathva
3,493630
3,493630
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
asked Apr 8 at 14:02
Utkarsh VermaUtkarsh Verma
1247
1247
$begingroup$
Please show your work solving using derivatives; doesn't $theta$ depend on $a$ and $b$?
$endgroup$
– J. W. Tanner
Apr 8 at 14:11
$begingroup$
@J.W.Tanner Done!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:28
$begingroup$
Thanks for clarifying; you have the correct answer using calculus now
$endgroup$
– J. W. Tanner
Apr 8 at 14:37
$begingroup$
Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
$endgroup$
– Jon
Apr 8 at 14:54
add a comment |
$begingroup$
Please show your work solving using derivatives; doesn't $theta$ depend on $a$ and $b$?
$endgroup$
– J. W. Tanner
Apr 8 at 14:11
$begingroup$
@J.W.Tanner Done!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:28
$begingroup$
Thanks for clarifying; you have the correct answer using calculus now
$endgroup$
– J. W. Tanner
Apr 8 at 14:37
$begingroup$
Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
$endgroup$
– Jon
Apr 8 at 14:54
$begingroup$
Please show your work solving using derivatives; doesn't $theta$ depend on $a$ and $b$?
$endgroup$
– J. W. Tanner
Apr 8 at 14:11
$begingroup$
Please show your work solving using derivatives; doesn't $theta$ depend on $a$ and $b$?
$endgroup$
– J. W. Tanner
Apr 8 at 14:11
$begingroup$
@J.W.Tanner Done!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:28
$begingroup$
@J.W.Tanner Done!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:28
$begingroup$
Thanks for clarifying; you have the correct answer using calculus now
$endgroup$
– J. W. Tanner
Apr 8 at 14:37
$begingroup$
Thanks for clarifying; you have the correct answer using calculus now
$endgroup$
– J. W. Tanner
Apr 8 at 14:37
$begingroup$
Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
$endgroup$
– Jon
Apr 8 at 14:54
$begingroup$
Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
$endgroup$
– Jon
Apr 8 at 14:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Even if $f(theta)ge g(theta)$ for all $theta$, $f$ may attain its least possible value for some $theta$ that doesn't minimize $g$. Consider for instance the inequality (which holds on the whole real line) $$theta^2ge frac12theta^2-theta-1$$ The LHS is minimized for $theta=0$, the RHS for $theta=1$.
So it's not really a matter of choice of approach. It's a matter of basic logic.
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
$endgroup$
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
add a comment |
$begingroup$
Let $y=asec t+bcsc t$
$dfracdydt=asec ttan t-bcsc tcot t=dfracasin^3t-bcos^3tcos^2tsin^2t $
For extreme values of $y,f'(t)=0$
$impliestan^3t=dfrac baiffdfrac acos^3t=dfrac bsin^3t=pmsqrta^2/3+b^2/3$
How have you found $t=dfracpi8?$
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
$endgroup$
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Even if $f(theta)ge g(theta)$ for all $theta$, $f$ may attain its least possible value for some $theta$ that doesn't minimize $g$. Consider for instance the inequality (which holds on the whole real line) $$theta^2ge frac12theta^2-theta-1$$ The LHS is minimized for $theta=0$, the RHS for $theta=1$.
So it's not really a matter of choice of approach. It's a matter of basic logic.
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
$endgroup$
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
add a comment |
$begingroup$
Even if $f(theta)ge g(theta)$ for all $theta$, $f$ may attain its least possible value for some $theta$ that doesn't minimize $g$. Consider for instance the inequality (which holds on the whole real line) $$theta^2ge frac12theta^2-theta-1$$ The LHS is minimized for $theta=0$, the RHS for $theta=1$.
So it's not really a matter of choice of approach. It's a matter of basic logic.
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
$endgroup$
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
add a comment |
$begingroup$
Even if $f(theta)ge g(theta)$ for all $theta$, $f$ may attain its least possible value for some $theta$ that doesn't minimize $g$. Consider for instance the inequality (which holds on the whole real line) $$theta^2ge frac12theta^2-theta-1$$ The LHS is minimized for $theta=0$, the RHS for $theta=1$.
So it's not really a matter of choice of approach. It's a matter of basic logic.
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
$endgroup$
Even if $f(theta)ge g(theta)$ for all $theta$, $f$ may attain its least possible value for some $theta$ that doesn't minimize $g$. Consider for instance the inequality (which holds on the whole real line) $$theta^2ge frac12theta^2-theta-1$$ The LHS is minimized for $theta=0$, the RHS for $theta=1$.
So it's not really a matter of choice of approach. It's a matter of basic logic.
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
edited Apr 8 at 14:21
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
answered Apr 8 at 14:12
Saucy O'PathSaucy O'Path
6,6301627
6,6301627
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
add a comment |
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:31
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
Do both and the one that works faster is the one you should have used first.
$endgroup$
– Saucy O'Path
Apr 8 at 14:32
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
$endgroup$
– Saucy O'Path
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
$begingroup$
Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:34
add a comment |
$begingroup$
Let $y=asec t+bcsc t$
$dfracdydt=asec ttan t-bcsc tcot t=dfracasin^3t-bcos^3tcos^2tsin^2t $
For extreme values of $y,f'(t)=0$
$impliestan^3t=dfrac baiffdfrac acos^3t=dfrac bsin^3t=pmsqrta^2/3+b^2/3$
How have you found $t=dfracpi8?$
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
$endgroup$
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
add a comment |
$begingroup$
Let $y=asec t+bcsc t$
$dfracdydt=asec ttan t-bcsc tcot t=dfracasin^3t-bcos^3tcos^2tsin^2t $
For extreme values of $y,f'(t)=0$
$impliestan^3t=dfrac baiffdfrac acos^3t=dfrac bsin^3t=pmsqrta^2/3+b^2/3$
How have you found $t=dfracpi8?$
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
$endgroup$
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
add a comment |
$begingroup$
Let $y=asec t+bcsc t$
$dfracdydt=asec ttan t-bcsc tcot t=dfracasin^3t-bcos^3tcos^2tsin^2t $
For extreme values of $y,f'(t)=0$
$impliestan^3t=dfrac baiffdfrac acos^3t=dfrac bsin^3t=pmsqrta^2/3+b^2/3$
How have you found $t=dfracpi8?$
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
$endgroup$
Let $y=asec t+bcsc t$
$dfracdydt=asec ttan t-bcsc tcot t=dfracasin^3t-bcos^3tcos^2tsin^2t $
For extreme values of $y,f'(t)=0$
$impliestan^3t=dfrac baiffdfrac acos^3t=dfrac bsin^3t=pmsqrta^2/3+b^2/3$
How have you found $t=dfracpi8?$
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
answered Apr 8 at 14:12
lab bhattacharjeelab bhattacharjee
228k15159279
228k15159279
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
add a comment |
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:29
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
@Utkarsh, So, put those values in $f'(t)$
$endgroup$
– lab bhattacharjee
Apr 8 at 14:38
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
$begingroup$
The answer comes $fracpi6$ which is different from the one given by AM-GM inequality.
$endgroup$
– Utkarsh Verma
Apr 8 at 14:40
add a comment |
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$begingroup$
Please show your work solving using derivatives; doesn't $theta$ depend on $a$ and $b$?
$endgroup$
– J. W. Tanner
Apr 8 at 14:11
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@J.W.Tanner Done!
$endgroup$
– Utkarsh Verma
Apr 8 at 14:28
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Thanks for clarifying; you have the correct answer using calculus now
$endgroup$
– J. W. Tanner
Apr 8 at 14:37
$begingroup$
Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
$endgroup$
– Jon
Apr 8 at 14:54