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Maxima & Minima
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Hi, how do I find the minimum value of f(x) = (4sin²x + 9cosec²x) using the concepts of maxima and minima? First I used the AM-GM inequality, I got 12, which is the right answer if we take a look at the graph of the function. Then I tried using the concepts of maxima/minima (which is supposed to work too).
If I differentiate it once and set it equal to zero, it gives me x = (2n+1)À/2, i.e odd multiples of À/2 (check out the attachment). Now if we put x = À/2, 3À/2, 5À/2, etc into the function, it gives me 13 as the answer, which is wrong. It's minimum value is 12. Where did it go wrong? Because concepts of maxima and minima should get us to the correct answer?
Thanks
calculus
asked Jul 30 at 5:06
À times e
10115
 |Â
show 1 more comment
up vote
2
down vote
favorite
Hi, how do I find the minimum value of f(x) = (4sin²x + 9cosec²x) using the concepts of maxima and minima? First I used the AM-GM inequality, I got 12, which is the right answer if we take a look at the graph of the function. Then I tried using the concepts of maxima/minima (which is supposed to work too).
If I differentiate it once and set it equal to zero, it gives me x = (2n+1)À/2, i.e odd multiples of À/2 (check out the attachment). Now if we put x = À/2, 3À/2, 5À/2, etc into the function, it gives me 13 as the answer, which is wrong. It's minimum value is 12. Where did it go wrong? Because concepts of maxima and minima should get us to the correct answer?
Thanks
calculus
asked Jul 30 at 5:06
À times e
10115
13 is correct. AM-GM is loose here.
– angryavian
Jul 30 at 5:13
If you take a look at the graph of this function, it's clear that it's minimum value is 12 and it goes as high as +infinity. How is 13 the correct answer then?
– Ã€ times e
Jul 30 at 5:16
Yeah, you're right. 13 is the correct answer. Actually I didn't read its graph correctly. My bad! Sorry. And thanks
– Ã€ times e
Jul 30 at 5:20
Instead of differentiating , you can write the function as : $(2sinx-3cosecx)^2 +12 $. Then you can conclude that the minimum value of $(2sinx-3cosecx)^2$ is $1$ and hence min value of $f(x)$ is 13 . Although this is not the best method.
– Alphanerd
Jul 30 at 5:36
How did you conclude that the minimum value of (2sinx - 3cosecx)² is 1? Can you please explain it?
– Ã€ times e
Jul 31 at 9:39
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Hi, how do I find the minimum value of f(x) = (4sin²x + 9cosec²x) using the concepts of maxima and minima? First I used the AM-GM inequality, I got 12, which is the right answer if we take a look at the graph of the function. Then I tried using the concepts of maxima/minima (which is supposed to work too).
If I differentiate it once and set it equal to zero, it gives me x = (2n+1)À/2, i.e odd multiples of À/2 (check out the attachment). Now if we put x = À/2, 3À/2, 5À/2, etc into the function, it gives me 13 as the answer, which is wrong. It's minimum value is 12. Where did it go wrong? Because concepts of maxima and minima should get us to the correct answer?
Thanks
calculus
asked Jul 30 at 5:06
À times e
10115
Hi, how do I find the minimum value of f(x) = (4sin²x + 9cosec²x) using the concepts of maxima and minima? First I used the AM-GM inequality, I got 12, which is the right answer if we take a look at the graph of the function. Then I tried using the concepts of maxima/minima (which is supposed to work too).
If I differentiate it once and set it equal to zero, it gives me x = (2n+1)À/2, i.e odd multiples of À/2 (check out the attachment). Now if we put x = À/2, 3À/2, 5À/2, etc into the function, it gives me 13 as the answer, which is wrong. It's minimum value is 12. Where did it go wrong? Because concepts of maxima and minima should get us to the correct answer?
Thanks
calculus
asked Jul 30 at 5:06
À times e
10115
asked Jul 30 at 5:06
À times e
10115
asked Jul 30 at 5:06
À times e
10115
asked Jul 30 at 5:06
À times e
10115
10115
13 is correct. AM-GM is loose here.
– angryavian
Jul 30 at 5:13
If you take a look at the graph of this function, it's clear that it's minimum value is 12 and it goes as high as +infinity. How is 13 the correct answer then?
– Ã€ times e
Jul 30 at 5:16
Yeah, you're right. 13 is the correct answer. Actually I didn't read its graph correctly. My bad! Sorry. And thanks
– Ã€ times e
Jul 30 at 5:20
Instead of differentiating , you can write the function as : $(2sinx-3cosecx)^2 +12 $. Then you can conclude that the minimum value of $(2sinx-3cosecx)^2$ is $1$ and hence min value of $f(x)$ is 13 . Although this is not the best method.
– Alphanerd
Jul 30 at 5:36
How did you conclude that the minimum value of (2sinx - 3cosecx)² is 1? Can you please explain it?
– Ã€ times e
Jul 31 at 9:39
 |Â
show 1 more comment
13 is correct. AM-GM is loose here.
– angryavian
Jul 30 at 5:13
If you take a look at the graph of this function, it's clear that it's minimum value is 12 and it goes as high as +infinity. How is 13 the correct answer then?
– Ã€ times e
Jul 30 at 5:16
Yeah, you're right. 13 is the correct answer. Actually I didn't read its graph correctly. My bad! Sorry. And thanks
– Ã€ times e
Jul 30 at 5:20
Instead of differentiating , you can write the function as : $(2sinx-3cosecx)^2 +12 $. Then you can conclude that the minimum value of $(2sinx-3cosecx)^2$ is $1$ and hence min value of $f(x)$ is 13 . Although this is not the best method.
– Alphanerd
Jul 30 at 5:36
How did you conclude that the minimum value of (2sinx - 3cosecx)² is 1? Can you please explain it?
– Ã€ times e
Jul 31 at 9:39
13 is correct. AM-GM is loose here.
– angryavian
Jul 30 at 5:13
13 is correct. AM-GM is loose here.
– angryavian
Jul 30 at 5:13
If you take a look at the graph of this function, it's clear that it's minimum value is 12 and it goes as high as +infinity. How is 13 the correct answer then?
– Ã€ times e
Jul 30 at 5:16
If you take a look at the graph of this function, it's clear that it's minimum value is 12 and it goes as high as +infinity. How is 13 the correct answer then?
– Ã€ times e
Jul 30 at 5:16
Yeah, you're right. 13 is the correct answer. Actually I didn't read its graph correctly. My bad! Sorry. And thanks
– Ã€ times e
Jul 30 at 5:20
Yeah, you're right. 13 is the correct answer. Actually I didn't read its graph correctly. My bad! Sorry. And thanks
– Ã€ times e
Jul 30 at 5:20
Instead of differentiating , you can write the function as : $(2sinx-3cosecx)^2 +12 $. Then you can conclude that the minimum value of $(2sinx-3cosecx)^2$ is $1$ and hence min value of $f(x)$ is 13 . Although this is not the best method.
– Alphanerd
Jul 30 at 5:36
Instead of differentiating , you can write the function as : $(2sinx-3cosecx)^2 +12 $. Then you can conclude that the minimum value of $(2sinx-3cosecx)^2$ is $1$ and hence min value of $f(x)$ is 13 . Although this is not the best method.
– Alphanerd
Jul 30 at 5:36
How did you conclude that the minimum value of (2sinx - 3cosecx)² is 1? Can you please explain it?
– Ã€ times e
Jul 31 at 9:39
How did you conclude that the minimum value of (2sinx - 3cosecx)² is 1? Can you please explain it?
– Ã€ times e
Jul 31 at 9:39
 |Â
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
2
down vote
Using AM-GM inequality,
$$4sin^2 x + 9 cosec^2 x ge 2 sqrt4cdot 9=12$$
However, the equality only holds when $$4 sin^2 x = 9 cosec^2 x$$ $$sin^4 x = frac94$$which is not attainable.
Desmos link here for the graph.
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
add a comment |Â
up vote
2
down vote
Solution
beginalign*4sin^2 x + 9 csc^2 x&=4sin^2 x + frac9 sin^2 x\&=left(4sin^2+ frac4 sin^2 xright)+frac5sin^2 x\&geq 2sqrt4sin^2 xcdotfrac4sin^2 x+5\&=2cdot 4+5\&=13,endalign*with the equality holding if and only if $sin^2 x=1$, namely, $x=2kpipm dfracpi2$ where $k in mathbbZ.$
answered Jul 30 at 9:55
mengdie1982
2,827216
add a comment |Â
up vote
1
down vote
Writing $sin^2x=t^2$ for $tin[0,1]$, we want to know when $g(t)=4t^2+frac9t^2$ becomes minimal. We have $4t^2+frac9t^2 = (frac3t-2t)^2+12$. Dropping the square gives you the 12. However, $frac3t-2tge 1$ for $0< tle 1$ and so the function must be at least 13, attained whenever $sin^2x=1$.
answered Jul 30 at 6:35
Kusma
1,027111
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
add a comment |Â
up vote
0
down vote
As noted we can't apply AM-GM inequality but we can refer to Rearrangement inequality selecting
- $(a_1,a_2)=left(sin^2 x,frac94 right)$
- $(b_1,b_2)=left(4,frac4sin^2 xright)$
we have
$$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$
$$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$
answered Jul 30 at 14:03
gimusi
64.5k73482
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Using AM-GM inequality,
$$4sin^2 x + 9 cosec^2 x ge 2 sqrt4cdot 9=12$$
However, the equality only holds when $$4 sin^2 x = 9 cosec^2 x$$ $$sin^4 x = frac94$$which is not attainable.
Desmos link here for the graph.
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
add a comment |Â
up vote
2
down vote
Using AM-GM inequality,
$$4sin^2 x + 9 cosec^2 x ge 2 sqrt4cdot 9=12$$
However, the equality only holds when $$4 sin^2 x = 9 cosec^2 x$$ $$sin^4 x = frac94$$which is not attainable.
Desmos link here for the graph.
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using AM-GM inequality,
$$4sin^2 x + 9 cosec^2 x ge 2 sqrt4cdot 9=12$$
However, the equality only holds when $$4 sin^2 x = 9 cosec^2 x$$ $$sin^4 x = frac94$$which is not attainable.
Desmos link here for the graph.
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
Using AM-GM inequality,
$$4sin^2 x + 9 cosec^2 x ge 2 sqrt4cdot 9=12$$
However, the equality only holds when $$4 sin^2 x = 9 cosec^2 x$$ $$sin^4 x = frac94$$which is not attainable.
Desmos link here for the graph.
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
answered Jul 30 at 5:17
Siong Thye Goh
76.8k134794
76.8k134794
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
add a comment |Â
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
Thanks a lot. Now I understand it
– Ã€ times e
Jul 30 at 5:21
add a comment |Â
up vote
2
down vote
Solution
beginalign*4sin^2 x + 9 csc^2 x&=4sin^2 x + frac9 sin^2 x\&=left(4sin^2+ frac4 sin^2 xright)+frac5sin^2 x\&geq 2sqrt4sin^2 xcdotfrac4sin^2 x+5\&=2cdot 4+5\&=13,endalign*with the equality holding if and only if $sin^2 x=1$, namely, $x=2kpipm dfracpi2$ where $k in mathbbZ.$
answered Jul 30 at 9:55
mengdie1982
2,827216
add a comment |Â
up vote
2
down vote
Solution
beginalign*4sin^2 x + 9 csc^2 x&=4sin^2 x + frac9 sin^2 x\&=left(4sin^2+ frac4 sin^2 xright)+frac5sin^2 x\&geq 2sqrt4sin^2 xcdotfrac4sin^2 x+5\&=2cdot 4+5\&=13,endalign*with the equality holding if and only if $sin^2 x=1$, namely, $x=2kpipm dfracpi2$ where $k in mathbbZ.$
answered Jul 30 at 9:55
mengdie1982
2,827216
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Solution
beginalign*4sin^2 x + 9 csc^2 x&=4sin^2 x + frac9 sin^2 x\&=left(4sin^2+ frac4 sin^2 xright)+frac5sin^2 x\&geq 2sqrt4sin^2 xcdotfrac4sin^2 x+5\&=2cdot 4+5\&=13,endalign*with the equality holding if and only if $sin^2 x=1$, namely, $x=2kpipm dfracpi2$ where $k in mathbbZ.$
answered Jul 30 at 9:55
mengdie1982
2,827216
Solution
beginalign*4sin^2 x + 9 csc^2 x&=4sin^2 x + frac9 sin^2 x\&=left(4sin^2+ frac4 sin^2 xright)+frac5sin^2 x\&geq 2sqrt4sin^2 xcdotfrac4sin^2 x+5\&=2cdot 4+5\&=13,endalign*with the equality holding if and only if $sin^2 x=1$, namely, $x=2kpipm dfracpi2$ where $k in mathbbZ.$
answered Jul 30 at 9:55
mengdie1982
2,827216
answered Jul 30 at 9:55
mengdie1982
2,827216
answered Jul 30 at 9:55
mengdie1982
2,827216
answered Jul 30 at 9:55
mengdie1982
2,827216
2,827216
add a comment |Â
add a comment |Â
up vote
1
down vote
Writing $sin^2x=t^2$ for $tin[0,1]$, we want to know when $g(t)=4t^2+frac9t^2$ becomes minimal. We have $4t^2+frac9t^2 = (frac3t-2t)^2+12$. Dropping the square gives you the 12. However, $frac3t-2tge 1$ for $0< tle 1$ and so the function must be at least 13, attained whenever $sin^2x=1$.
answered Jul 30 at 6:35
Kusma
1,027111
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
add a comment |Â
up vote
1
down vote
Writing $sin^2x=t^2$ for $tin[0,1]$, we want to know when $g(t)=4t^2+frac9t^2$ becomes minimal. We have $4t^2+frac9t^2 = (frac3t-2t)^2+12$. Dropping the square gives you the 12. However, $frac3t-2tge 1$ for $0< tle 1$ and so the function must be at least 13, attained whenever $sin^2x=1$.
answered Jul 30 at 6:35
Kusma
1,027111
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Writing $sin^2x=t^2$ for $tin[0,1]$, we want to know when $g(t)=4t^2+frac9t^2$ becomes minimal. We have $4t^2+frac9t^2 = (frac3t-2t)^2+12$. Dropping the square gives you the 12. However, $frac3t-2tge 1$ for $0< tle 1$ and so the function must be at least 13, attained whenever $sin^2x=1$.
answered Jul 30 at 6:35
Kusma
1,027111
Writing $sin^2x=t^2$ for $tin[0,1]$, we want to know when $g(t)=4t^2+frac9t^2$ becomes minimal. We have $4t^2+frac9t^2 = (frac3t-2t)^2+12$. Dropping the square gives you the 12. However, $frac3t-2tge 1$ for $0< tle 1$ and so the function must be at least 13, attained whenever $sin^2x=1$.
answered Jul 30 at 6:35
Kusma
1,027111
answered Jul 30 at 6:35
Kusma
1,027111
answered Jul 30 at 6:35
Kusma
1,027111
answered Jul 30 at 6:35
Kusma
1,027111
1,027111
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
add a comment |Â
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
How to establish the last inequality?
– lab bhattacharjee
Jul 30 at 8:49
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
Calculus works, for example. The derivative of $frac3t-2t$ is negative for $t>0$, so the function is $ge$ its value at the rightmost point. But the answer of @mengdie1982 is more elegant anyway :)
– Kusma
Jul 30 at 10:06
add a comment |Â
up vote
0
down vote
As noted we can't apply AM-GM inequality but we can refer to Rearrangement inequality selecting
- $(a_1,a_2)=left(sin^2 x,frac94 right)$
- $(b_1,b_2)=left(4,frac4sin^2 xright)$
we have
$$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$
$$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$
answered Jul 30 at 14:03
gimusi
64.5k73482
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
add a comment |Â
up vote
0
down vote
As noted we can't apply AM-GM inequality but we can refer to Rearrangement inequality selecting
- $(a_1,a_2)=left(sin^2 x,frac94 right)$
- $(b_1,b_2)=left(4,frac4sin^2 xright)$
we have
$$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$
$$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$
answered Jul 30 at 14:03
gimusi
64.5k73482
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As noted we can't apply AM-GM inequality but we can refer to Rearrangement inequality selecting
- $(a_1,a_2)=left(sin^2 x,frac94 right)$
- $(b_1,b_2)=left(4,frac4sin^2 xright)$
we have
$$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$
$$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$
answered Jul 30 at 14:03
gimusi
64.5k73482
As noted we can't apply AM-GM inequality but we can refer to Rearrangement inequality selecting
- $(a_1,a_2)=left(sin^2 x,frac94 right)$
- $(b_1,b_2)=left(4,frac4sin^2 xright)$
we have
$$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$
$$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$
answered Jul 30 at 14:03
gimusi
64.5k73482
edited Jul 30 at 15:04
answered Jul 30 at 14:03
gimusi
64.5k73482
answered Jul 30 at 14:03
gimusi
64.5k73482
answered Jul 30 at 14:03
gimusi
64.5k73482
64.5k73482
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
add a comment |Â
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
this is wrong，you can not do like this，unless you can go on with the sencond inequality.
– mengdie1982
Jul 30 at 14:33
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
@mengdie1982 Yes of course you are right! I fix, thanks
– gimusi
Jul 30 at 14:56
add a comment |Â
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13 is correct. AM-GM is loose here.
– angryavian
Jul 30 at 5:13
If you take a look at the graph of this function, it's clear that it's minimum value is 12 and it goes as high as +infinity. How is 13 the correct answer then?
– Ã€ times e
Jul 30 at 5:16
Yeah, you're right. 13 is the correct answer. Actually I didn't read its graph correctly. My bad! Sorry. And thanks
– Ã€ times e
Jul 30 at 5:20
Instead of differentiating , you can write the function as : $(2sinx-3cosecx)^2 +12 $. Then you can conclude that the minimum value of $(2sinx-3cosecx)^2$ is $1$ and hence min value of $f(x)$ is 13 . Although this is not the best method.
– Alphanerd
Jul 30 at 5:36
How did you conclude that the minimum value of (2sinx - 3cosecx)² is 1? Can you please explain it?
– Ã€ times e
Jul 31 at 9:39