AS Differentiation












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stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.



"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.



So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$



This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.










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  • 1




    $begingroup$
    Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
    $endgroup$
    – bounceback
    Jan 7 at 20:36


















0












$begingroup$


stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.



"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.



So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$



This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
    $endgroup$
    – bounceback
    Jan 7 at 20:36
















0












0








0





$begingroup$


stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.



"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.



So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$



This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.










share|cite|improve this question









$endgroup$




stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.



"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.



So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$



This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.







derivatives






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asked Jan 7 at 20:32









Jia Xuan NgJia Xuan Ng

101




101








  • 1




    $begingroup$
    Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
    $endgroup$
    – bounceback
    Jan 7 at 20:36
















  • 1




    $begingroup$
    Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
    $endgroup$
    – bounceback
    Jan 7 at 20:36










1




1




$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36






$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36












2 Answers
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$begingroup$

$$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$



From here, you correctly solved for $h$ in terms of $r$:



$$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$



You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.



$$V = bh = pi r^2h$$



Plugging in $(1)$, you get



$$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$



Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:



$$frac{dV}{dr} = 25pi-3pi r^2$$



$$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$



Can you continue from here?



Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
    $endgroup$
    – Jia Xuan Ng
    Jan 7 at 22:41










  • $begingroup$
    You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
    $endgroup$
    – KM101
    Jan 7 at 22:55












  • $begingroup$
    I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
    $endgroup$
    – Jia Xuan Ng
    Jan 8 at 18:42












  • $begingroup$
    Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
    $endgroup$
    – KM101
    Jan 8 at 18:53












  • $begingroup$
    Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
    $endgroup$
    – Jia Xuan Ng
    Jan 8 at 18:58



















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$begingroup$

Hint:
The volume of a Cylinder $V=r^2pi h$
Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$



Now look for the maximum of that Function






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    2 Answers
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    active

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    2 Answers
    2






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    active

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    0












    $begingroup$

    $$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$



    From here, you correctly solved for $h$ in terms of $r$:



    $$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$



    You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.



    $$V = bh = pi r^2h$$



    Plugging in $(1)$, you get



    $$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$



    Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:



    $$frac{dV}{dr} = 25pi-3pi r^2$$



    $$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$



    Can you continue from here?



    Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
      $endgroup$
      – Jia Xuan Ng
      Jan 7 at 22:41










    • $begingroup$
      You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
      $endgroup$
      – KM101
      Jan 7 at 22:55












    • $begingroup$
      I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:42












    • $begingroup$
      Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
      $endgroup$
      – KM101
      Jan 8 at 18:53












    • $begingroup$
      Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:58
















    0












    $begingroup$

    $$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$



    From here, you correctly solved for $h$ in terms of $r$:



    $$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$



    You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.



    $$V = bh = pi r^2h$$



    Plugging in $(1)$, you get



    $$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$



    Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:



    $$frac{dV}{dr} = 25pi-3pi r^2$$



    $$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$



    Can you continue from here?



    Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
      $endgroup$
      – Jia Xuan Ng
      Jan 7 at 22:41










    • $begingroup$
      You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
      $endgroup$
      – KM101
      Jan 7 at 22:55












    • $begingroup$
      I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:42












    • $begingroup$
      Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
      $endgroup$
      – KM101
      Jan 8 at 18:53












    • $begingroup$
      Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:58














    0












    0








    0





    $begingroup$

    $$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$



    From here, you correctly solved for $h$ in terms of $r$:



    $$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$



    You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.



    $$V = bh = pi r^2h$$



    Plugging in $(1)$, you get



    $$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$



    Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:



    $$frac{dV}{dr} = 25pi-3pi r^2$$



    $$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$



    Can you continue from here?



    Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.






    share|cite|improve this answer











    $endgroup$



    $$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$



    From here, you correctly solved for $h$ in terms of $r$:



    $$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$



    You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.



    $$V = bh = pi r^2h$$



    Plugging in $(1)$, you get



    $$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$



    Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:



    $$frac{dV}{dr} = 25pi-3pi r^2$$



    $$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$



    Can you continue from here?



    Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 7 at 22:57

























    answered Jan 7 at 21:00









    KM101KM101

    6,0251525




    6,0251525












    • $begingroup$
      I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
      $endgroup$
      – Jia Xuan Ng
      Jan 7 at 22:41










    • $begingroup$
      You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
      $endgroup$
      – KM101
      Jan 7 at 22:55












    • $begingroup$
      I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:42












    • $begingroup$
      Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
      $endgroup$
      – KM101
      Jan 8 at 18:53












    • $begingroup$
      Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:58


















    • $begingroup$
      I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
      $endgroup$
      – Jia Xuan Ng
      Jan 7 at 22:41










    • $begingroup$
      You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
      $endgroup$
      – KM101
      Jan 7 at 22:55












    • $begingroup$
      I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:42












    • $begingroup$
      Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
      $endgroup$
      – KM101
      Jan 8 at 18:53












    • $begingroup$
      Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
      $endgroup$
      – Jia Xuan Ng
      Jan 8 at 18:58
















    $begingroup$
    I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
    $endgroup$
    – Jia Xuan Ng
    Jan 7 at 22:41




    $begingroup$
    I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
    $endgroup$
    – Jia Xuan Ng
    Jan 7 at 22:41












    $begingroup$
    You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
    $endgroup$
    – KM101
    Jan 7 at 22:55






    $begingroup$
    You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
    $endgroup$
    – KM101
    Jan 7 at 22:55














    $begingroup$
    I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
    $endgroup$
    – Jia Xuan Ng
    Jan 8 at 18:42






    $begingroup$
    I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
    $endgroup$
    – Jia Xuan Ng
    Jan 8 at 18:42














    $begingroup$
    Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
    $endgroup$
    – KM101
    Jan 8 at 18:53






    $begingroup$
    Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
    $endgroup$
    – KM101
    Jan 8 at 18:53














    $begingroup$
    Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
    $endgroup$
    – Jia Xuan Ng
    Jan 8 at 18:58




    $begingroup$
    Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
    $endgroup$
    – Jia Xuan Ng
    Jan 8 at 18:58











    0












    $begingroup$

    Hint:
    The volume of a Cylinder $V=r^2pi h$
    Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$



    Now look for the maximum of that Function






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint:
      The volume of a Cylinder $V=r^2pi h$
      Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$



      Now look for the maximum of that Function






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:
        The volume of a Cylinder $V=r^2pi h$
        Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$



        Now look for the maximum of that Function






        share|cite|improve this answer









        $endgroup$



        Hint:
        The volume of a Cylinder $V=r^2pi h$
        Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$



        Now look for the maximum of that Function







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 20:42









        A. PA. P

        1085




        1085






























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