Given $a=r^2pi h,c=2rpi h+r^2pi$, how to find $r$ and $h$ in terms of $a$ and $c$?












0












$begingroup$



Given the values of $a$ and $c$ and that
$$a=r^2pi h,quad
c=2rpi h+r^2pi,$$

what are the values of $r$ and $h$?




So recently on a math test, I got this question which I couldn't seem to figure out. It was really boggling my mind and I later found out that I just had interpreted the question wrong, but I was still set on finding out the solution to the problem. I can't seem to do it and wonder if anyone could help.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You only need the first equation. $$r=sqrt{frac a{pi h}}$$
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:57






  • 1




    $begingroup$
    How did you interpret the problem? What do a, and c represent as variables? It looks like $a$ is the volume of a cylinder with radius r and height h, and $c$ is the surface area of the open cylinder: bottom + sides.
    $endgroup$
    – Namaste
    Dec 18 '18 at 20:58








  • 2




    $begingroup$
    But @DonThousand, $h$ is unknown as well. So we need two equations to solve for $r$, $h$.
    $endgroup$
    – Namaste
    Dec 18 '18 at 21:00










  • $begingroup$
    @DonThousand has shown the answer to the question in the title. To also find $h$, plug that expression into the second equation, that should give you a cubic equation in $sqrt h$. Those can be solved.
    $endgroup$
    – Henrik
    Dec 18 '18 at 21:01












  • $begingroup$
    Doesn't seem pleasant. From the first, $h=frac a{pi r^2}$ and if you substitute that into the second you are led to $pi r^3-cr+2a=0$, so you have a cubic to deal with.
    $endgroup$
    – lulu
    Dec 18 '18 at 21:02
















0












$begingroup$



Given the values of $a$ and $c$ and that
$$a=r^2pi h,quad
c=2rpi h+r^2pi,$$

what are the values of $r$ and $h$?




So recently on a math test, I got this question which I couldn't seem to figure out. It was really boggling my mind and I later found out that I just had interpreted the question wrong, but I was still set on finding out the solution to the problem. I can't seem to do it and wonder if anyone could help.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You only need the first equation. $$r=sqrt{frac a{pi h}}$$
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:57






  • 1




    $begingroup$
    How did you interpret the problem? What do a, and c represent as variables? It looks like $a$ is the volume of a cylinder with radius r and height h, and $c$ is the surface area of the open cylinder: bottom + sides.
    $endgroup$
    – Namaste
    Dec 18 '18 at 20:58








  • 2




    $begingroup$
    But @DonThousand, $h$ is unknown as well. So we need two equations to solve for $r$, $h$.
    $endgroup$
    – Namaste
    Dec 18 '18 at 21:00










  • $begingroup$
    @DonThousand has shown the answer to the question in the title. To also find $h$, plug that expression into the second equation, that should give you a cubic equation in $sqrt h$. Those can be solved.
    $endgroup$
    – Henrik
    Dec 18 '18 at 21:01












  • $begingroup$
    Doesn't seem pleasant. From the first, $h=frac a{pi r^2}$ and if you substitute that into the second you are led to $pi r^3-cr+2a=0$, so you have a cubic to deal with.
    $endgroup$
    – lulu
    Dec 18 '18 at 21:02














0












0








0





$begingroup$



Given the values of $a$ and $c$ and that
$$a=r^2pi h,quad
c=2rpi h+r^2pi,$$

what are the values of $r$ and $h$?




So recently on a math test, I got this question which I couldn't seem to figure out. It was really boggling my mind and I later found out that I just had interpreted the question wrong, but I was still set on finding out the solution to the problem. I can't seem to do it and wonder if anyone could help.






algebra-precalculus







share|cite|improve this question











$endgroup$





Given the values of $a$ and $c$ and that
$$a=r^2pi h,quad
c=2rpi h+r^2pi,$$

what are the values of $r$ and $h$?




So recently on a math test, I got this question which I couldn't seem to figure out. It was really boggling my mind and I later found out that I just had interpreted the question wrong, but I was still set on finding out the solution to the problem. I can't seem to do it and wonder if anyone could help.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 23:42







user587192

















asked Dec 18 '18 at 20:54









HelloHello

1




1












  • $begingroup$
    You only need the first equation. $$r=sqrt{frac a{pi h}}$$
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:57






  • 1




    $begingroup$
    How did you interpret the problem? What do a, and c represent as variables? It looks like $a$ is the volume of a cylinder with radius r and height h, and $c$ is the surface area of the open cylinder: bottom + sides.
    $endgroup$
    – Namaste
    Dec 18 '18 at 20:58








  • 2




    $begingroup$
    But @DonThousand, $h$ is unknown as well. So we need two equations to solve for $r$, $h$.
    $endgroup$
    – Namaste
    Dec 18 '18 at 21:00










  • $begingroup$
    @DonThousand has shown the answer to the question in the title. To also find $h$, plug that expression into the second equation, that should give you a cubic equation in $sqrt h$. Those can be solved.
    $endgroup$
    – Henrik
    Dec 18 '18 at 21:01












  • $begingroup$
    Doesn't seem pleasant. From the first, $h=frac a{pi r^2}$ and if you substitute that into the second you are led to $pi r^3-cr+2a=0$, so you have a cubic to deal with.
    $endgroup$
    – lulu
    Dec 18 '18 at 21:02


















  • $begingroup$
    You only need the first equation. $$r=sqrt{frac a{pi h}}$$
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:57






  • 1




    $begingroup$
    How did you interpret the problem? What do a, and c represent as variables? It looks like $a$ is the volume of a cylinder with radius r and height h, and $c$ is the surface area of the open cylinder: bottom + sides.
    $endgroup$
    – Namaste
    Dec 18 '18 at 20:58








  • 2




    $begingroup$
    But @DonThousand, $h$ is unknown as well. So we need two equations to solve for $r$, $h$.
    $endgroup$
    – Namaste
    Dec 18 '18 at 21:00










  • $begingroup$
    @DonThousand has shown the answer to the question in the title. To also find $h$, plug that expression into the second equation, that should give you a cubic equation in $sqrt h$. Those can be solved.
    $endgroup$
    – Henrik
    Dec 18 '18 at 21:01












  • $begingroup$
    Doesn't seem pleasant. From the first, $h=frac a{pi r^2}$ and if you substitute that into the second you are led to $pi r^3-cr+2a=0$, so you have a cubic to deal with.
    $endgroup$
    – lulu
    Dec 18 '18 at 21:02
















$begingroup$
You only need the first equation. $$r=sqrt{frac a{pi h}}$$
$endgroup$
– Don Thousand
Dec 18 '18 at 20:57




$begingroup$
You only need the first equation. $$r=sqrt{frac a{pi h}}$$
$endgroup$
– Don Thousand
Dec 18 '18 at 20:57




1




1




$begingroup$
How did you interpret the problem? What do a, and c represent as variables? It looks like $a$ is the volume of a cylinder with radius r and height h, and $c$ is the surface area of the open cylinder: bottom + sides.
$endgroup$
– Namaste
Dec 18 '18 at 20:58






$begingroup$
How did you interpret the problem? What do a, and c represent as variables? It looks like $a$ is the volume of a cylinder with radius r and height h, and $c$ is the surface area of the open cylinder: bottom + sides.
$endgroup$
– Namaste
Dec 18 '18 at 20:58






2




2




$begingroup$
But @DonThousand, $h$ is unknown as well. So we need two equations to solve for $r$, $h$.
$endgroup$
– Namaste
Dec 18 '18 at 21:00




$begingroup$
But @DonThousand, $h$ is unknown as well. So we need two equations to solve for $r$, $h$.
$endgroup$
– Namaste
Dec 18 '18 at 21:00












$begingroup$
@DonThousand has shown the answer to the question in the title. To also find $h$, plug that expression into the second equation, that should give you a cubic equation in $sqrt h$. Those can be solved.
$endgroup$
– Henrik
Dec 18 '18 at 21:01






$begingroup$
@DonThousand has shown the answer to the question in the title. To also find $h$, plug that expression into the second equation, that should give you a cubic equation in $sqrt h$. Those can be solved.
$endgroup$
– Henrik
Dec 18 '18 at 21:01














$begingroup$
Doesn't seem pleasant. From the first, $h=frac a{pi r^2}$ and if you substitute that into the second you are led to $pi r^3-cr+2a=0$, so you have a cubic to deal with.
$endgroup$
– lulu
Dec 18 '18 at 21:02




$begingroup$
Doesn't seem pleasant. From the first, $h=frac a{pi r^2}$ and if you substitute that into the second you are led to $pi r^3-cr+2a=0$, so you have a cubic to deal with.
$endgroup$
– lulu
Dec 18 '18 at 21:02










1 Answer
1






active

oldest

votes


















-1












$begingroup$

From the first equation we have that
$$
2pi r h = 2frac{a}{r}
$$
so we can substitute to the second one and then
$$
c=2frac{a}{r}+r^2 piqquadtext{so}qquad cr=2a+r^3pi
$$
and now the only thing you have to do is to solve this equation using Cardano formula and get that
$$
r=frac{left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } aright)^{2/3}+sqrt[3]{3}
c}{3^{2/3} sqrt{pi } sqrt[3]{sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } a}}
$$
and now by substituting to the first formula you get
$$
h=frac{a}{r^2 pi}=frac{3 sqrt[3]{3} a left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}}{left(left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}+sqrt[3]{3} cright)^2}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:06








  • 1




    $begingroup$
    @amWhy now you have the result in answer
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:14












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

From the first equation we have that
$$
2pi r h = 2frac{a}{r}
$$
so we can substitute to the second one and then
$$
c=2frac{a}{r}+r^2 piqquadtext{so}qquad cr=2a+r^3pi
$$
and now the only thing you have to do is to solve this equation using Cardano formula and get that
$$
r=frac{left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } aright)^{2/3}+sqrt[3]{3}
c}{3^{2/3} sqrt{pi } sqrt[3]{sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } a}}
$$
and now by substituting to the first formula you get
$$
h=frac{a}{r^2 pi}=frac{3 sqrt[3]{3} a left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}}{left(left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}+sqrt[3]{3} cright)^2}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:06








  • 1




    $begingroup$
    @amWhy now you have the result in answer
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:14
















-1












$begingroup$

From the first equation we have that
$$
2pi r h = 2frac{a}{r}
$$
so we can substitute to the second one and then
$$
c=2frac{a}{r}+r^2 piqquadtext{so}qquad cr=2a+r^3pi
$$
and now the only thing you have to do is to solve this equation using Cardano formula and get that
$$
r=frac{left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } aright)^{2/3}+sqrt[3]{3}
c}{3^{2/3} sqrt{pi } sqrt[3]{sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } a}}
$$
and now by substituting to the first formula you get
$$
h=frac{a}{r^2 pi}=frac{3 sqrt[3]{3} a left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}}{left(left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}+sqrt[3]{3} cright)^2}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:06








  • 1




    $begingroup$
    @amWhy now you have the result in answer
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:14














-1












-1








-1





$begingroup$

From the first equation we have that
$$
2pi r h = 2frac{a}{r}
$$
so we can substitute to the second one and then
$$
c=2frac{a}{r}+r^2 piqquadtext{so}qquad cr=2a+r^3pi
$$
and now the only thing you have to do is to solve this equation using Cardano formula and get that
$$
r=frac{left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } aright)^{2/3}+sqrt[3]{3}
c}{3^{2/3} sqrt{pi } sqrt[3]{sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } a}}
$$
and now by substituting to the first formula you get
$$
h=frac{a}{r^2 pi}=frac{3 sqrt[3]{3} a left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}}{left(left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}+sqrt[3]{3} cright)^2}
$$






share|cite|improve this answer











$endgroup$



From the first equation we have that
$$
2pi r h = 2frac{a}{r}
$$
so we can substitute to the second one and then
$$
c=2frac{a}{r}+r^2 piqquadtext{so}qquad cr=2a+r^3pi
$$
and now the only thing you have to do is to solve this equation using Cardano formula and get that
$$
r=frac{left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } aright)^{2/3}+sqrt[3]{3}
c}{3^{2/3} sqrt{pi } sqrt[3]{sqrt{81 pi a^2-3 c^3}-9 sqrt{pi } a}}
$$
and now by substituting to the first formula you get
$$
h=frac{a}{r^2 pi}=frac{3 sqrt[3]{3} a left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}}{left(left(sqrt{81 pi a^2-3 c^3}-9 sqrt{pi }
aright)^{2/3}+sqrt[3]{3} cright)^2}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 21:13

























answered Dec 18 '18 at 21:04









avan1235avan1235

3688




3688












  • $begingroup$
    so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:06








  • 1




    $begingroup$
    @amWhy now you have the result in answer
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:14


















  • $begingroup$
    so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:06








  • 1




    $begingroup$
    @amWhy now you have the result in answer
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:14
















$begingroup$
so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
$endgroup$
– avan1235
Dec 18 '18 at 21:06






$begingroup$
so using the Cardano formula you get the solution for r and then you substitute r to the first equation and get the $h$ value
$endgroup$
– avan1235
Dec 18 '18 at 21:06






1




1




$begingroup$
@amWhy now you have the result in answer
$endgroup$
– avan1235
Dec 18 '18 at 21:14




$begingroup$
@amWhy now you have the result in answer
$endgroup$
– avan1235
Dec 18 '18 at 21:14


















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