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How to calculate the smallest surface needed to have a cylinder of 1.75 dm^3?
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I'm new to optimization, and I'm doing simple exercises. Nevertheless I have the impression that, in order to find the smallest surface needed to get a $1.75$ dm^3 cylinder, I should look for the minima of a function $f_s(r,h)= 2pi r h + 2pi r^2$ such that $pi r^2h=1.75$, which implies going MUCH further than my current level by working on a bivariate function.
What is the simplest way of solving this problem?
Thanks in advance for your answer.
calculus optimization
asked Jul 27 at 18:07
torito verdejo
374
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up vote
1
down vote
favorite
I'm new to optimization, and I'm doing simple exercises. Nevertheless I have the impression that, in order to find the smallest surface needed to get a $1.75$ dm^3 cylinder, I should look for the minima of a function $f_s(r,h)= 2pi r h + 2pi r^2$ such that $pi r^2h=1.75$, which implies going MUCH further than my current level by working on a bivariate function.
What is the simplest way of solving this problem?
Thanks in advance for your answer.
calculus optimization
asked Jul 27 at 18:07
torito verdejo
374
1
I believe it should be $f_s(r,h)=2pi r(h+r)$ which is the total surface area of a cylinder, otherwise as $rtoinfty$ you'd get $textSurface Areato0$.
– MalayTheDynamo
Jul 27 at 18:14
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm new to optimization, and I'm doing simple exercises. Nevertheless I have the impression that, in order to find the smallest surface needed to get a $1.75$ dm^3 cylinder, I should look for the minima of a function $f_s(r,h)= 2pi r h + 2pi r^2$ such that $pi r^2h=1.75$, which implies going MUCH further than my current level by working on a bivariate function.
What is the simplest way of solving this problem?
Thanks in advance for your answer.
calculus optimization
asked Jul 27 at 18:07
torito verdejo
374
I'm new to optimization, and I'm doing simple exercises. Nevertheless I have the impression that, in order to find the smallest surface needed to get a $1.75$ dm^3 cylinder, I should look for the minima of a function $f_s(r,h)= 2pi r h + 2pi r^2$ such that $pi r^2h=1.75$, which implies going MUCH further than my current level by working on a bivariate function.
What is the simplest way of solving this problem?
Thanks in advance for your answer.
calculus optimization
asked Jul 27 at 18:07
torito verdejo
374
edited Jul 27 at 18:22
asked Jul 27 at 18:07
torito verdejo
374
asked Jul 27 at 18:07
torito verdejo
374
asked Jul 27 at 18:07
torito verdejo
374
374
1
I believe it should be $f_s(r,h)=2pi r(h+r)$ which is the total surface area of a cylinder, otherwise as $rtoinfty$ you'd get $textSurface Areato0$.
– MalayTheDynamo
Jul 27 at 18:14
add a comment |Â
1
I believe it should be $f_s(r,h)=2pi r(h+r)$ which is the total surface area of a cylinder, otherwise as $rtoinfty$ you'd get $textSurface Areato0$.
– MalayTheDynamo
Jul 27 at 18:14
1
1
I believe it should be $f_s(r,h)=2pi r(h+r)$ which is the total surface area of a cylinder, otherwise as $rtoinfty$ you'd get $textSurface Areato0$.
– MalayTheDynamo
Jul 27 at 18:14
I believe it should be $f_s(r,h)=2pi r(h+r)$ which is the total surface area of a cylinder, otherwise as $rtoinfty$ you'd get $textSurface Areato0$.
– MalayTheDynamo
Jul 27 at 18:14
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
HINT
We have
- $pi r^2h=1.75 implies pi rh=frac1.75r$
and then including the bases
- $f_s(r)= 2pi r(r + h)=2pi r^2+frac3.5r$
answered Jul 27 at 18:17
gimusi
64.9k73483
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up vote
1
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$$pi r^2 h = 1.75$$
$$h = frac1.75pi r^2$$
What you want to optimize is
$$2pi r (h+r) = 2pi r cdot left(frac1.75pi r^2+rright)=frac3.5r+2pi r^2$$ subject to $r >0$.
Now the question is single variable and hopefully you can take it from here.
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
add a comment |Â
up vote
1
down vote
It is given the Surface of this cylinder:
$$A=2pi r^2+2pi rh$$ and the volume $$V=7/4=pi r^2h$$.
Solve this equation for $$h$$ and plugg it in the Surface Formula, which containes then only one variable.
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
HINT
We have
- $pi r^2h=1.75 implies pi rh=frac1.75r$
and then including the bases
- $f_s(r)= 2pi r(r + h)=2pi r^2+frac3.5r$
answered Jul 27 at 18:17
gimusi
64.9k73483
add a comment |Â
up vote
1
down vote
accepted
HINT
We have
- $pi r^2h=1.75 implies pi rh=frac1.75r$
and then including the bases
- $f_s(r)= 2pi r(r + h)=2pi r^2+frac3.5r$
answered Jul 27 at 18:17
gimusi
64.9k73483
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
HINT
We have
- $pi r^2h=1.75 implies pi rh=frac1.75r$
and then including the bases
- $f_s(r)= 2pi r(r + h)=2pi r^2+frac3.5r$
answered Jul 27 at 18:17
gimusi
64.9k73483
HINT
We have
- $pi r^2h=1.75 implies pi rh=frac1.75r$
and then including the bases
- $f_s(r)= 2pi r(r + h)=2pi r^2+frac3.5r$
answered Jul 27 at 18:17
gimusi
64.9k73483
answered Jul 27 at 18:17
gimusi
64.9k73483
answered Jul 27 at 18:17
gimusi
64.9k73483
answered Jul 27 at 18:17
gimusi
64.9k73483
64.9k73483
add a comment |Â
add a comment |Â
up vote
1
down vote
$$pi r^2 h = 1.75$$
$$h = frac1.75pi r^2$$
What you want to optimize is
$$2pi r (h+r) = 2pi r cdot left(frac1.75pi r^2+rright)=frac3.5r+2pi r^2$$ subject to $r >0$.
Now the question is single variable and hopefully you can take it from here.
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
add a comment |Â
up vote
1
down vote
$$pi r^2 h = 1.75$$
$$h = frac1.75pi r^2$$
What you want to optimize is
$$2pi r (h+r) = 2pi r cdot left(frac1.75pi r^2+rright)=frac3.5r+2pi r^2$$ subject to $r >0$.
Now the question is single variable and hopefully you can take it from here.
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$pi r^2 h = 1.75$$
$$h = frac1.75pi r^2$$
What you want to optimize is
$$2pi r (h+r) = 2pi r cdot left(frac1.75pi r^2+rright)=frac3.5r+2pi r^2$$ subject to $r >0$.
Now the question is single variable and hopefully you can take it from here.
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
$$pi r^2 h = 1.75$$
$$h = frac1.75pi r^2$$
What you want to optimize is
$$2pi r (h+r) = 2pi r cdot left(frac1.75pi r^2+rright)=frac3.5r+2pi r^2$$ subject to $r >0$.
Now the question is single variable and hopefully you can take it from here.
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
answered Jul 27 at 18:18
Siong Thye Goh
77k134794
77k134794
add a comment |Â
add a comment |Â
up vote
1
down vote
It is given the Surface of this cylinder:
$$A=2pi r^2+2pi rh$$ and the volume $$V=7/4=pi r^2h$$.
Solve this equation for $$h$$ and plugg it in the Surface Formula, which containes then only one variable.
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
1
down vote
It is given the Surface of this cylinder:
$$A=2pi r^2+2pi rh$$ and the volume $$V=7/4=pi r^2h$$.
Solve this equation for $$h$$ and plugg it in the Surface Formula, which containes then only one variable.
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is given the Surface of this cylinder:
$$A=2pi r^2+2pi rh$$ and the volume $$V=7/4=pi r^2h$$.
Solve this equation for $$h$$ and plugg it in the Surface Formula, which containes then only one variable.
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
It is given the Surface of this cylinder:
$$A=2pi r^2+2pi rh$$ and the volume $$V=7/4=pi r^2h$$.
Solve this equation for $$h$$ and plugg it in the Surface Formula, which containes then only one variable.
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
answered Jul 27 at 18:18
Dr. Sonnhard Graubner
66.7k32659
66.7k32659
add a comment |Â
add a comment |Â
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1
I believe it should be $f_s(r,h)=2pi r(h+r)$ which is the total surface area of a cylinder, otherwise as $rtoinfty$ you'd get $textSurface Areato0$.
– MalayTheDynamo
Jul 27 at 18:14