Height and Radius for maximum volume cylinder of given surface area
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I've got a school project, in which I'm supposed to get a radius of a base and a height of a cylinder while having the maximal volume possible. Surface area of the cylinder is known.
I found a great solution on this forum (https://math.stackexchange.com/a/1449593/621973), but the problem is that I need to explain my teacher, how I turned $$frac k{(k+1)^{3/2}}$$ from V(k) into $$f(k)=lnfrac{k}{(k+1)^{3/2}}$$ By the way, I can't give her another solution so I really need to find explanation for this one.
I'll be glad for every idea.
P.S.: Sorry for my english.
calculus geometry
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
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add a comment |
$begingroup$
I've got a school project, in which I'm supposed to get a radius of a base and a height of a cylinder while having the maximal volume possible. Surface area of the cylinder is known.
I found a great solution on this forum (https://math.stackexchange.com/a/1449593/621973), but the problem is that I need to explain my teacher, how I turned $$frac k{(k+1)^{3/2}}$$ from V(k) into $$f(k)=lnfrac{k}{(k+1)^{3/2}}$$ By the way, I can't give her another solution so I really need to find explanation for this one.
I'll be glad for every idea.
P.S.: Sorry for my english.
calculus geometry
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
$endgroup$
add a comment |
$begingroup$
I've got a school project, in which I'm supposed to get a radius of a base and a height of a cylinder while having the maximal volume possible. Surface area of the cylinder is known.
I found a great solution on this forum (https://math.stackexchange.com/a/1449593/621973), but the problem is that I need to explain my teacher, how I turned $$frac k{(k+1)^{3/2}}$$ from V(k) into $$f(k)=lnfrac{k}{(k+1)^{3/2}}$$ By the way, I can't give her another solution so I really need to find explanation for this one.
I'll be glad for every idea.
P.S.: Sorry for my english.
calculus geometry
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
$endgroup$
I've got a school project, in which I'm supposed to get a radius of a base and a height of a cylinder while having the maximal volume possible. Surface area of the cylinder is known.
I found a great solution on this forum (https://math.stackexchange.com/a/1449593/621973), but the problem is that I need to explain my teacher, how I turned $$frac k{(k+1)^{3/2}}$$ from V(k) into $$f(k)=lnfrac{k}{(k+1)^{3/2}}$$ By the way, I can't give her another solution so I really need to find explanation for this one.
I'll be glad for every idea.
P.S.: Sorry for my english.
calculus geometry
calculus geometry
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
asked Dec 2 '18 at 10:31
Adam S.Adam S.
1
1
add a comment |
add a comment |
1 Answer
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You can maximize
$$f(k)=frac k{(k+1)^{3/2}}$$ or its logarithm
$$g(k)=log left(frac{k}{(k+1)^{3/2}}right)=log(k)-frac 32 log(k+1)$$ $g(k)$ is a better choice since its derivative is easier to get.
This is also the principle of logarithmic differentiation.
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You can maximize
$$f(k)=frac k{(k+1)^{3/2}}$$ or its logarithm
$$g(k)=log left(frac{k}{(k+1)^{3/2}}right)=log(k)-frac 32 log(k+1)$$ $g(k)$ is a better choice since its derivative is easier to get.
This is also the principle of logarithmic differentiation.
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
add a comment |
$begingroup$
You can maximize
$$f(k)=frac k{(k+1)^{3/2}}$$ or its logarithm
$$g(k)=log left(frac{k}{(k+1)^{3/2}}right)=log(k)-frac 32 log(k+1)$$ $g(k)$ is a better choice since its derivative is easier to get.
This is also the principle of logarithmic differentiation.
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
add a comment |
$begingroup$
You can maximize
$$f(k)=frac k{(k+1)^{3/2}}$$ or its logarithm
$$g(k)=log left(frac{k}{(k+1)^{3/2}}right)=log(k)-frac 32 log(k+1)$$ $g(k)$ is a better choice since its derivative is easier to get.
This is also the principle of logarithmic differentiation.
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
You can maximize
$$f(k)=frac k{(k+1)^{3/2}}$$ or its logarithm
$$g(k)=log left(frac{k}{(k+1)^{3/2}}right)=log(k)-frac 32 log(k+1)$$ $g(k)$ is a better choice since its derivative is easier to get.
This is also the principle of logarithmic differentiation.
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
answered Dec 2 '18 at 10:44
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
add a comment |
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
$begingroup$
I thought it's some method due to which it is used like that. So the point of this step is just to make it easier for the following steps?
$endgroup$
– Adam S.
Dec 2 '18 at 11:01
add a comment |
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