Minimising the surface area of a Cuboid with a different length, width, and height.
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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
asked Jan 2 at 11:00
hhalawehhhalaweh
32
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Jan 2 at 11:00
hhalawehhhalaweh
32
$endgroup$
$begingroup$
"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
|
show 1 more comment
$begingroup$
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
asked Jan 2 at 11:00
hhalawehhhalaweh
32
$endgroup$
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
asked Jan 2 at 11:00
hhalawehhhalaweh
32
asked Jan 2 at 11:00
hhalawehhhalaweh
32
edited Jan 2 at 11:06
hhalaweh
asked Jan 2 at 11:00
hhalawehhhalaweh
32
asked Jan 2 at 11:00
hhalawehhhalaweh
32
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
|
show 1 more comment
$begingroup$
"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
|
show 1 more comment
1 Answer
1
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oldest
votes
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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 $6sqrt[3]{V^2}.$
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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 $6sqrt[3]{V^2}.$
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $6sqrt[3]{V^2}.$
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $6sqrt[3]{V^2}.$
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
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 $6sqrt[3]{V^2}.$
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 2 at 11:35
Michael RozenbergMichael Rozenberg
106k1893198
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
"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