If $a+b=1$ find the greatest value for $a^2b^3$












4












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I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated






algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality







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  • 4




    $begingroup$
    I presume you want $a$, $bge0$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:14










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    13 hours ago
















4












$begingroup$


I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated






algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I presume you want $a$, $bge0$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:14










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    13 hours ago














4












4








4


2



$begingroup$


I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated






algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$




I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated






algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality




algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality






share|cite|improve this question















share|cite|improve this question













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edited Jan 5 at 6:32









Michael Rozenberg

97.6k1589188




97.6k1589188










asked Jan 5 at 6:12









user587054user587054

46011




46011








  • 4




    $begingroup$
    I presume you want $a$, $bge0$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:14










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    13 hours ago














  • 4




    $begingroup$
    I presume you want $a$, $bge0$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:14










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    13 hours ago








4




4




$begingroup$
I presume you want $a$, $bge0$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 6:14




$begingroup$
I presume you want $a$, $bge0$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 6:14












$begingroup$
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.
$endgroup$
– Carl Mummert
13 hours ago




$begingroup$
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.
$endgroup$
– Carl Mummert
13 hours ago










5 Answers
5






active

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8












$begingroup$

Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
$$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
etc.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    Just using algebra.



    Using $b=1-a$, you are looking for the maximum of function
    $$f(a)=a^2(1-a)^3$$ Compute the derivatives
    $$f'(a)=-a(1-a)^2 (5 a-2)$$
    $$f''(a)=2(1-a)(10 a^2-8 a+1)$$
    The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Although it's not the most elegant, this is by far the most straightforward answer!
      $endgroup$
      – Toby Mak
      Jan 5 at 6:55










    • $begingroup$
      You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
      $endgroup$
      – Toby Mak
      Jan 5 at 7:02










    • $begingroup$
      Is $f''(x)$ required at all?
      $endgroup$
      – Shubham Johri
      Jan 5 at 7:03






    • 2




      $begingroup$
      @ShubhamJohri; Just to confirm that this is a maximum.
      $endgroup$
      – Claude Leibovici
      Jan 5 at 7:08



















    5












    $begingroup$

    Does not exist. Try $brightarrow+infty$.



    For non-negatives $a$ and $b$ by AM-GM we obtain:
    $$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
    The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      $$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
      $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
      $$f'(x) = 0 implies x in {0, 1, frac 35}$$
      $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
      $$f''(0)=0, f''(1)>0, f''(frac 35)<0$$



      Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
        $endgroup$
        – Mark Bennet
        Jan 5 at 6:57










      • $begingroup$
        We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
        $endgroup$
        – Shubham Johri
        Jan 5 at 7:06





















      1












      $begingroup$

      Let Constraint function be
      $$C(a,b)= a+b-1=0 $$
      and the Object function that needs maximization be
      $$G(a,b)=a^2b^3 $$
      we have with partial differentiation of combined Lagrangian:
      $$ C(a,b)- lambda G(a,b) $$
      condition to evaluate the multiplier $lambda$
      $$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
      $$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$



      That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.



      After plugging in these values Object function maximum value is:



      $$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$






      share|cite|improve this answer









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        5 Answers
        5






        active

        oldest

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        5 Answers
        5






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

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        8












        $begingroup$

        Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
        $$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
        etc.






        share|cite|improve this answer









        $endgroup$


















          8












          $begingroup$

          Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
          $$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
          etc.






          share|cite|improve this answer









          $endgroup$
















            8












            8








            8





            $begingroup$

            Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
            $$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
            etc.






            share|cite|improve this answer









            $endgroup$



            Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
            $$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
            etc.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 6:16









            Lord Shark the UnknownLord Shark the Unknown

            102k959132




            102k959132























                7












                $begingroup$

                Just using algebra.



                Using $b=1-a$, you are looking for the maximum of function
                $$f(a)=a^2(1-a)^3$$ Compute the derivatives
                $$f'(a)=-a(1-a)^2 (5 a-2)$$
                $$f''(a)=2(1-a)(10 a^2-8 a+1)$$
                The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Although it's not the most elegant, this is by far the most straightforward answer!
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 6:55










                • $begingroup$
                  You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 7:02










                • $begingroup$
                  Is $f''(x)$ required at all?
                  $endgroup$
                  – Shubham Johri
                  Jan 5 at 7:03






                • 2




                  $begingroup$
                  @ShubhamJohri; Just to confirm that this is a maximum.
                  $endgroup$
                  – Claude Leibovici
                  Jan 5 at 7:08
















                7












                $begingroup$

                Just using algebra.



                Using $b=1-a$, you are looking for the maximum of function
                $$f(a)=a^2(1-a)^3$$ Compute the derivatives
                $$f'(a)=-a(1-a)^2 (5 a-2)$$
                $$f''(a)=2(1-a)(10 a^2-8 a+1)$$
                The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Although it's not the most elegant, this is by far the most straightforward answer!
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 6:55










                • $begingroup$
                  You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 7:02










                • $begingroup$
                  Is $f''(x)$ required at all?
                  $endgroup$
                  – Shubham Johri
                  Jan 5 at 7:03






                • 2




                  $begingroup$
                  @ShubhamJohri; Just to confirm that this is a maximum.
                  $endgroup$
                  – Claude Leibovici
                  Jan 5 at 7:08














                7












                7








                7





                $begingroup$

                Just using algebra.



                Using $b=1-a$, you are looking for the maximum of function
                $$f(a)=a^2(1-a)^3$$ Compute the derivatives
                $$f'(a)=-a(1-a)^2 (5 a-2)$$
                $$f''(a)=2(1-a)(10 a^2-8 a+1)$$
                The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.






                share|cite|improve this answer









                $endgroup$



                Just using algebra.



                Using $b=1-a$, you are looking for the maximum of function
                $$f(a)=a^2(1-a)^3$$ Compute the derivatives
                $$f'(a)=-a(1-a)^2 (5 a-2)$$
                $$f''(a)=2(1-a)(10 a^2-8 a+1)$$
                The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 6:52









                Claude LeiboviciClaude Leibovici

                119k1157132




                119k1157132












                • $begingroup$
                  Although it's not the most elegant, this is by far the most straightforward answer!
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 6:55










                • $begingroup$
                  You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 7:02










                • $begingroup$
                  Is $f''(x)$ required at all?
                  $endgroup$
                  – Shubham Johri
                  Jan 5 at 7:03






                • 2




                  $begingroup$
                  @ShubhamJohri; Just to confirm that this is a maximum.
                  $endgroup$
                  – Claude Leibovici
                  Jan 5 at 7:08


















                • $begingroup$
                  Although it's not the most elegant, this is by far the most straightforward answer!
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 6:55










                • $begingroup$
                  You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
                  $endgroup$
                  – Toby Mak
                  Jan 5 at 7:02










                • $begingroup$
                  Is $f''(x)$ required at all?
                  $endgroup$
                  – Shubham Johri
                  Jan 5 at 7:03






                • 2




                  $begingroup$
                  @ShubhamJohri; Just to confirm that this is a maximum.
                  $endgroup$
                  – Claude Leibovici
                  Jan 5 at 7:08
















                $begingroup$
                Although it's not the most elegant, this is by far the most straightforward answer!
                $endgroup$
                – Toby Mak
                Jan 5 at 6:55




                $begingroup$
                Although it's not the most elegant, this is by far the most straightforward answer!
                $endgroup$
                – Toby Mak
                Jan 5 at 6:55












                $begingroup$
                You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
                $endgroup$
                – Toby Mak
                Jan 5 at 7:02




                $begingroup$
                You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
                $endgroup$
                – Toby Mak
                Jan 5 at 7:02












                $begingroup$
                Is $f''(x)$ required at all?
                $endgroup$
                – Shubham Johri
                Jan 5 at 7:03




                $begingroup$
                Is $f''(x)$ required at all?
                $endgroup$
                – Shubham Johri
                Jan 5 at 7:03




                2




                2




                $begingroup$
                @ShubhamJohri; Just to confirm that this is a maximum.
                $endgroup$
                – Claude Leibovici
                Jan 5 at 7:08




                $begingroup$
                @ShubhamJohri; Just to confirm that this is a maximum.
                $endgroup$
                – Claude Leibovici
                Jan 5 at 7:08











                5












                $begingroup$

                Does not exist. Try $brightarrow+infty$.



                For non-negatives $a$ and $b$ by AM-GM we obtain:
                $$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
                The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.






                share|cite|improve this answer











                $endgroup$


















                  5












                  $begingroup$

                  Does not exist. Try $brightarrow+infty$.



                  For non-negatives $a$ and $b$ by AM-GM we obtain:
                  $$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
                  The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.






                  share|cite|improve this answer











                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Does not exist. Try $brightarrow+infty$.



                    For non-negatives $a$ and $b$ by AM-GM we obtain:
                    $$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
                    The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.






                    share|cite|improve this answer











                    $endgroup$



                    Does not exist. Try $brightarrow+infty$.



                    For non-negatives $a$ and $b$ by AM-GM we obtain:
                    $$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
                    The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 5 at 6:33

























                    answered Jan 5 at 6:21









                    Michael RozenbergMichael Rozenberg

                    97.6k1589188




                    97.6k1589188























                        1












                        $begingroup$

                        $$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
                        $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
                        $$f'(x) = 0 implies x in {0, 1, frac 35}$$
                        $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
                        $$f''(0)=0, f''(1)>0, f''(frac 35)<0$$



                        Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
                          $endgroup$
                          – Mark Bennet
                          Jan 5 at 6:57










                        • $begingroup$
                          We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
                          $endgroup$
                          – Shubham Johri
                          Jan 5 at 7:06


















                        1












                        $begingroup$

                        $$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
                        $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
                        $$f'(x) = 0 implies x in {0, 1, frac 35}$$
                        $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
                        $$f''(0)=0, f''(1)>0, f''(frac 35)<0$$



                        Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
                          $endgroup$
                          – Mark Bennet
                          Jan 5 at 6:57










                        • $begingroup$
                          We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
                          $endgroup$
                          – Shubham Johri
                          Jan 5 at 7:06
















                        1












                        1








                        1





                        $begingroup$

                        $$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
                        $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
                        $$f'(x) = 0 implies x in {0, 1, frac 35}$$
                        $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
                        $$f''(0)=0, f''(1)>0, f''(frac 35)<0$$



                        Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$






                        share|cite|improve this answer









                        $endgroup$



                        $$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
                        $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
                        $$f'(x) = 0 implies x in {0, 1, frac 35}$$
                        $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
                        $$f''(0)=0, f''(1)>0, f''(frac 35)<0$$



                        Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 5 at 6:48









                        Rhys HughesRhys Hughes

                        5,1501427




                        5,1501427












                        • $begingroup$
                          I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
                          $endgroup$
                          – Mark Bennet
                          Jan 5 at 6:57










                        • $begingroup$
                          We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
                          $endgroup$
                          – Shubham Johri
                          Jan 5 at 7:06




















                        • $begingroup$
                          I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
                          $endgroup$
                          – Mark Bennet
                          Jan 5 at 6:57










                        • $begingroup$
                          We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
                          $endgroup$
                          – Shubham Johri
                          Jan 5 at 7:06


















                        $begingroup$
                        I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
                        $endgroup$
                        – Mark Bennet
                        Jan 5 at 6:57




                        $begingroup$
                        I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
                        $endgroup$
                        – Mark Bennet
                        Jan 5 at 6:57












                        $begingroup$
                        We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
                        $endgroup$
                        – Shubham Johri
                        Jan 5 at 7:06






                        $begingroup$
                        We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
                        $endgroup$
                        – Shubham Johri
                        Jan 5 at 7:06













                        1












                        $begingroup$

                        Let Constraint function be
                        $$C(a,b)= a+b-1=0 $$
                        and the Object function that needs maximization be
                        $$G(a,b)=a^2b^3 $$
                        we have with partial differentiation of combined Lagrangian:
                        $$ C(a,b)- lambda G(a,b) $$
                        condition to evaluate the multiplier $lambda$
                        $$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
                        $$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$



                        That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.



                        After plugging in these values Object function maximum value is:



                        $$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Let Constraint function be
                          $$C(a,b)= a+b-1=0 $$
                          and the Object function that needs maximization be
                          $$G(a,b)=a^2b^3 $$
                          we have with partial differentiation of combined Lagrangian:
                          $$ C(a,b)- lambda G(a,b) $$
                          condition to evaluate the multiplier $lambda$
                          $$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
                          $$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$



                          That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.



                          After plugging in these values Object function maximum value is:



                          $$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let Constraint function be
                            $$C(a,b)= a+b-1=0 $$
                            and the Object function that needs maximization be
                            $$G(a,b)=a^2b^3 $$
                            we have with partial differentiation of combined Lagrangian:
                            $$ C(a,b)- lambda G(a,b) $$
                            condition to evaluate the multiplier $lambda$
                            $$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
                            $$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$



                            That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.



                            After plugging in these values Object function maximum value is:



                            $$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$






                            share|cite|improve this answer









                            $endgroup$



                            Let Constraint function be
                            $$C(a,b)= a+b-1=0 $$
                            and the Object function that needs maximization be
                            $$G(a,b)=a^2b^3 $$
                            we have with partial differentiation of combined Lagrangian:
                            $$ C(a,b)- lambda G(a,b) $$
                            condition to evaluate the multiplier $lambda$
                            $$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
                            $$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$



                            That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.



                            After plugging in these values Object function maximum value is:



                            $$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 8:26









                            NarasimhamNarasimham

                            20.6k52158




                            20.6k52158






























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