Minimising the surface area of a Cuboid with a different length, width, and height.












-2












$begingroup$


I've been trying to minimise the surface of a Cuboid, with a different length, width, and height, but I haven't been able to do so, considering that there is more than 2 variables.



The constraint being the volume of the cuboid.



Considering the following equations:



S.A of cuboid = 2(wl+hl+hw)
V = whl



Any help would be appreciated.



Thanks!






calculus optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Minimise" subject to what constraints?? If there are no constraints then the $0 times 0 times 0$ cuboid would have the smallest surface area
    $endgroup$
    – glowstonetrees
    Jan 2 at 11:02










  • $begingroup$
    @glowstonetrees the constraint would be the volume of the Cuboid.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:03










  • $begingroup$
    You can set the variables to be the length $l$ and width $w$, and the height is known because volume is constant. Therefore, you only have two variables. Then look at the area of the cuboid, $A=A(l,w)$. You need to find the values of length and width so that that function gets the minimum value.
    $endgroup$
    – Matti P.
    Jan 2 at 11:06












  • $begingroup$
    @MattiP. With two variables still remaining unknown, it wouldn't be possible to find the first derivative.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:22










  • $begingroup$
    @hhalaweh Say me please.Is the volum a constant?
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 11:29
















-2












$begingroup$


I've been trying to minimise the surface of a Cuboid, with a different length, width, and height, but I haven't been able to do so, considering that there is more than 2 variables.



The constraint being the volume of the cuboid.



Considering the following equations:



S.A of cuboid = 2(wl+hl+hw)
V = whl



Any help would be appreciated.



Thanks!






calculus optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Minimise" subject to what constraints?? If there are no constraints then the $0 times 0 times 0$ cuboid would have the smallest surface area
    $endgroup$
    – glowstonetrees
    Jan 2 at 11:02










  • $begingroup$
    @glowstonetrees the constraint would be the volume of the Cuboid.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:03










  • $begingroup$
    You can set the variables to be the length $l$ and width $w$, and the height is known because volume is constant. Therefore, you only have two variables. Then look at the area of the cuboid, $A=A(l,w)$. You need to find the values of length and width so that that function gets the minimum value.
    $endgroup$
    – Matti P.
    Jan 2 at 11:06












  • $begingroup$
    @MattiP. With two variables still remaining unknown, it wouldn't be possible to find the first derivative.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:22










  • $begingroup$
    @hhalaweh Say me please.Is the volum a constant?
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 11:29














-2












-2








-2





$begingroup$


I've been trying to minimise the surface of a Cuboid, with a different length, width, and height, but I haven't been able to do so, considering that there is more than 2 variables.



The constraint being the volume of the cuboid.



Considering the following equations:



S.A of cuboid = 2(wl+hl+hw)
V = whl



Any help would be appreciated.



Thanks!






calculus optimization







share|cite|improve this question











$endgroup$




I've been trying to minimise the surface of a Cuboid, with a different length, width, and height, but I haven't been able to do so, considering that there is more than 2 variables.



The constraint being the volume of the cuboid.



Considering the following equations:



S.A of cuboid = 2(wl+hl+hw)
V = whl



Any help would be appreciated.



Thanks!






calculus optimization




calculus optimization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 11:06







hhalaweh

















asked Jan 2 at 11:00









hhalawehhhalaweh

32




32












  • $begingroup$
    "Minimise" subject to what constraints?? If there are no constraints then the $0 times 0 times 0$ cuboid would have the smallest surface area
    $endgroup$
    – glowstonetrees
    Jan 2 at 11:02










  • $begingroup$
    @glowstonetrees the constraint would be the volume of the Cuboid.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:03










  • $begingroup$
    You can set the variables to be the length $l$ and width $w$, and the height is known because volume is constant. Therefore, you only have two variables. Then look at the area of the cuboid, $A=A(l,w)$. You need to find the values of length and width so that that function gets the minimum value.
    $endgroup$
    – Matti P.
    Jan 2 at 11:06












  • $begingroup$
    @MattiP. With two variables still remaining unknown, it wouldn't be possible to find the first derivative.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:22










  • $begingroup$
    @hhalaweh Say me please.Is the volum a constant?
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 11:29


















  • $begingroup$
    "Minimise" subject to what constraints?? If there are no constraints then the $0 times 0 times 0$ cuboid would have the smallest surface area
    $endgroup$
    – glowstonetrees
    Jan 2 at 11:02










  • $begingroup$
    @glowstonetrees the constraint would be the volume of the Cuboid.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:03










  • $begingroup$
    You can set the variables to be the length $l$ and width $w$, and the height is known because volume is constant. Therefore, you only have two variables. Then look at the area of the cuboid, $A=A(l,w)$. You need to find the values of length and width so that that function gets the minimum value.
    $endgroup$
    – Matti P.
    Jan 2 at 11:06












  • $begingroup$
    @MattiP. With two variables still remaining unknown, it wouldn't be possible to find the first derivative.
    $endgroup$
    – hhalaweh
    Jan 2 at 11:22










  • $begingroup$
    @hhalaweh Say me please.Is the volum a constant?
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 11:29
















$begingroup$
"Minimise" subject to what constraints?? If there are no constraints then the $0 times 0 times 0$ cuboid would have the smallest surface area
$endgroup$
– glowstonetrees
Jan 2 at 11:02




$begingroup$
"Minimise" subject to what constraints?? If there are no constraints then the $0 times 0 times 0$ cuboid would have the smallest surface area
$endgroup$
– glowstonetrees
Jan 2 at 11:02












$begingroup$
@glowstonetrees the constraint would be the volume of the Cuboid.
$endgroup$
– hhalaweh
Jan 2 at 11:03




$begingroup$
@glowstonetrees the constraint would be the volume of the Cuboid.
$endgroup$
– hhalaweh
Jan 2 at 11:03












$begingroup$
You can set the variables to be the length $l$ and width $w$, and the height is known because volume is constant. Therefore, you only have two variables. Then look at the area of the cuboid, $A=A(l,w)$. You need to find the values of length and width so that that function gets the minimum value.
$endgroup$
– Matti P.
Jan 2 at 11:06






$begingroup$
You can set the variables to be the length $l$ and width $w$, and the height is known because volume is constant. Therefore, you only have two variables. Then look at the area of the cuboid, $A=A(l,w)$. You need to find the values of length and width so that that function gets the minimum value.
$endgroup$
– Matti P.
Jan 2 at 11:06














$begingroup$
@MattiP. With two variables still remaining unknown, it wouldn't be possible to find the first derivative.
$endgroup$
– hhalaweh
Jan 2 at 11:22




$begingroup$
@MattiP. With two variables still remaining unknown, it wouldn't be possible to find the first derivative.
$endgroup$
– hhalaweh
Jan 2 at 11:22












$begingroup$
@hhalaweh Say me please.Is the volum a constant?
$endgroup$
– Michael Rozenberg
Jan 2 at 11:29




$begingroup$
@hhalaweh Say me please.Is the volum a constant?
$endgroup$
– Michael Rozenberg
Jan 2 at 11:29










1 Answer
1






active

oldest

votes


















0












$begingroup$

If it means that the volum $V$ is given we can make the following.



By AM-GM
$$2(wl+hl+hw)geq2cdot3sqrt[3]{wlcdot hlcdot hw}=6sqrt[3]{w^2l^2h^2}=6sqrt[3]{V^2}.$$



The equality occurs for $w=l=h,$ which is impossible by the given.



Id est, the minimal value does not exist, but the infimum is $6sqrt[3]{V^2}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can I use the infimum to find the minimum surface area?
    $endgroup$
    – hhalaweh
    Jan 2 at 11:40










  • $begingroup$
    @hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 12:09











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

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0












$begingroup$

If it means that the volum $V$ is given we can make the following.



By AM-GM
$$2(wl+hl+hw)geq2cdot3sqrt[3]{wlcdot hlcdot hw}=6sqrt[3]{w^2l^2h^2}=6sqrt[3]{V^2}.$$



The equality occurs for $w=l=h,$ which is impossible by the given.



Id est, the minimal value does not exist, but the infimum is $6sqrt[3]{V^2}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can I use the infimum to find the minimum surface area?
    $endgroup$
    – hhalaweh
    Jan 2 at 11:40










  • $begingroup$
    @hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 12:09
















0












$begingroup$

If it means that the volum $V$ is given we can make the following.



By AM-GM
$$2(wl+hl+hw)geq2cdot3sqrt[3]{wlcdot hlcdot hw}=6sqrt[3]{w^2l^2h^2}=6sqrt[3]{V^2}.$$



The equality occurs for $w=l=h,$ which is impossible by the given.



Id est, the minimal value does not exist, but the infimum is $6sqrt[3]{V^2}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can I use the infimum to find the minimum surface area?
    $endgroup$
    – hhalaweh
    Jan 2 at 11:40










  • $begingroup$
    @hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 12:09














0












0








0





$begingroup$

If it means that the volum $V$ is given we can make the following.



By AM-GM
$$2(wl+hl+hw)geq2cdot3sqrt[3]{wlcdot hlcdot hw}=6sqrt[3]{w^2l^2h^2}=6sqrt[3]{V^2}.$$



The equality occurs for $w=l=h,$ which is impossible by the given.



Id est, the minimal value does not exist, but the infimum is $6sqrt[3]{V^2}.$






share|cite|improve this answer









$endgroup$



If it means that the volum $V$ is given we can make the following.



By AM-GM
$$2(wl+hl+hw)geq2cdot3sqrt[3]{wlcdot hlcdot hw}=6sqrt[3]{w^2l^2h^2}=6sqrt[3]{V^2}.$$



The equality occurs for $w=l=h,$ which is impossible by the given.



Id est, the minimal value does not exist, but the infimum is $6sqrt[3]{V^2}.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 11:35









Michael RozenbergMichael Rozenberg

106k1893198




106k1893198












  • $begingroup$
    can I use the infimum to find the minimum surface area?
    $endgroup$
    – hhalaweh
    Jan 2 at 11:40










  • $begingroup$
    @hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 12:09


















  • $begingroup$
    can I use the infimum to find the minimum surface area?
    $endgroup$
    – hhalaweh
    Jan 2 at 11:40










  • $begingroup$
    @hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 12:09
















$begingroup$
can I use the infimum to find the minimum surface area?
$endgroup$
– hhalaweh
Jan 2 at 11:40




$begingroup$
can I use the infimum to find the minimum surface area?
$endgroup$
– hhalaweh
Jan 2 at 11:40












$begingroup$
@hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:09




$begingroup$
@hhalaweh No, because the minimum does not occur. If $w$, $l$ and $h$ can be equal then the infimum is equal to the minimum.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:09


















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