Exercise XV num 12 - Calculus Made Easy

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A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.



Find its dimensions in order that the least amount of iron sheet may be used in its construction.



My approach:



Lets say all dimensions are each $ > 0$;



Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;



Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;



$frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;



$frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;



Is this the right path to continue? Looks too messy in my eyes.






calculus optimization partial-derivative







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    1














    A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.



    Find its dimensions in order that the least amount of iron sheet may be used in its construction.



    My approach:



    Lets say all dimensions are each $ > 0$;



    Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;



    Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;



    $frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;



    $frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;



    Is this the right path to continue? Looks too messy in my eyes.






    calculus optimization partial-derivative







    share|cite|improve this question









    New contributor




    LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      1












      1








      1







      A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.



      Find its dimensions in order that the least amount of iron sheet may be used in its construction.



      My approach:



      Lets say all dimensions are each $ > 0$;



      Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;



      Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;



      $frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;



      $frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;



      Is this the right path to continue? Looks too messy in my eyes.






      calculus optimization partial-derivative







      share|cite|improve this question









      New contributor




      LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.



      Find its dimensions in order that the least amount of iron sheet may be used in its construction.



      My approach:



      Lets say all dimensions are each $ > 0$;



      Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;



      Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;



      $frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;



      $frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;



      Is this the right path to continue? Looks too messy in my eyes.






      calculus optimization partial-derivative




      calculus optimization partial-derivative






      share|cite|improve this question









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      LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









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      asked 2 days ago









      LeoBonhartLeoBonhart

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          It is not so messy. Use $w=lambda h$ to get
          $$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
          $$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
          Subtract one from the other to get
          $$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.






          share|cite|improve this answer





















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            It is not so messy. Use $w=lambda h$ to get
            $$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
            $$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
            Subtract one from the other to get
            $$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.






            share|cite|improve this answer


























              0














              It is not so messy. Use $w=lambda h$ to get
              $$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
              $$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
              Subtract one from the other to get
              $$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.






              share|cite|improve this answer
























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                0






                It is not so messy. Use $w=lambda h$ to get
                $$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
                $$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
                Subtract one from the other to get
                $$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.






                share|cite|improve this answer












                It is not so messy. Use $w=lambda h$ to get
                $$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
                $$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
                Subtract one from the other to get
                $$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Claude LeiboviciClaude Leibovici

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                119k1157132






















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