Maxima & Minima

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
1












enter image description here



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)π/2, i.e odd multiples of π/2 (check out the attachment). Now if we put x = π/2, 3π/2, 5π/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




share|cite|improve this question



















  • 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?
    – Ï€ 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
    – Ï€ 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?
    – Ï€ times e
    Jul 31 at 9:39














up vote
2
down vote

favorite
1












enter image description here



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)π/2, i.e odd multiples of π/2 (check out the attachment). Now if we put x = π/2, 3π/2, 5π/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




share|cite|improve this question



















  • 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?
    – Ï€ 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
    – Ï€ 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?
    – Ï€ times e
    Jul 31 at 9:39












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





enter image description here



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)π/2, i.e odd multiples of π/2 (check out the attachment). Now if we put x = π/2, 3π/2, 5π/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




share|cite|improve this question











enter image description here



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)π/2, i.e odd multiples of π/2 (check out the attachment). Now if we put x = π/2, 3π/2, 5π/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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 30 at 5:06









π 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?
    – Ï€ 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
    – Ï€ 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?
    – Ï€ times e
    Jul 31 at 9:39
















  • 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?
    – Ï€ 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
    – Ï€ 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?
    – Ï€ 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?
– Ï€ 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?
– Ï€ 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
– Ï€ 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
– Ï€ 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?
– Ï€ 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?
– Ï€ times e
Jul 31 at 9:39










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.



enter image description here






share|cite|improve this answer





















  • Thanks a lot. Now I understand it
    – Ï€ times e
    Jul 30 at 5:21

















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 $k in mathbbZ.$






share|cite|improve this answer




























    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 $sin^2x=1$.






    share|cite|improve this answer





















    • 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

















    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



    $$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$



    $$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$






    share|cite|improve this answer























    • 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










    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866664%2fmaxima-minima%23new-answer', 'question_page');

    );

    Post as a guest

















    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.



    enter image description here






    share|cite|improve this answer





















    • Thanks a lot. Now I understand it
      – Ï€ times e
      Jul 30 at 5:21














    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.



    enter image description here






    share|cite|improve this answer





















    • Thanks a lot. Now I understand it
      – Ï€ times e
      Jul 30 at 5:21












    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.



    enter image description here






    share|cite|improve this answer













    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.



    enter image description here







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 30 at 5:17









    Siong Thye Goh

    76.8k134794




    76.8k134794











    • Thanks a lot. Now I understand it
      – Ï€ times e
      Jul 30 at 5:21
















    • Thanks a lot. Now I understand it
      – Ï€ times e
      Jul 30 at 5:21















    Thanks a lot. Now I understand it
    – Ï€ times e
    Jul 30 at 5:21




    Thanks a lot. Now I understand it
    – Ï€ times e
    Jul 30 at 5:21










    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 $k in mathbbZ.$






    share|cite|improve this answer

























      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 $k in mathbbZ.$






      share|cite|improve this answer























        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 $k in mathbbZ.$






        share|cite|improve this answer













        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 $k in mathbbZ.$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 30 at 9:55









        mengdie1982

        2,827216




        2,827216




















            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 $sin^2x=1$.






            share|cite|improve this answer





















            • 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














            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 $sin^2x=1$.






            share|cite|improve this answer





















            • 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












            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 $sin^2x=1$.






            share|cite|improve this answer













            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 $sin^2x=1$.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            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
















            • 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










            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



            $$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$



            $$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$






            share|cite|improve this answer























            • 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














            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



            $$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$



            $$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$






            share|cite|improve this answer























            • 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












            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



            $$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$



            $$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$






            share|cite|improve this answer















            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



            $$4sin^2 x + frac9 sin^2 x=4cdot sin^2 x + frac 94cdotfrac4 sin^2 x=$$



            $$=a_1b_1+a_2b_2ge a_1b_2+a_2b_1=sin^2cdot frac4 sin^2 x+ frac 94cdot4=13$$







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 30 at 15:04


























            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
















            • 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












             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866664%2fmaxima-minima%23new-answer', 'question_page');

            );

            Post as a guest
































































            Comments

            Post a Comment

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
            Read more

            Relationship between determinant of matrix and determinant of adjoint?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
            Read more

            Color the edges and diagonals of a regular polygon

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
            Read more