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Possible mistake finding the maximum volume of a box with the AM-GM inequality?
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I have found the following problem:
What is the box (without a top) of largest volume which can be constructed from
a square piece of tin of edge length $2a$ by cutting a square from each corner
and folding up the edges
I have tried to solve it with the AM-GM inequality: I have the sides $(2a-2h), (2a-2h),h$ and then:
$$frac(2a-2h)+ (2a-2h) + h3geq sqrt[3](2a-2h)^2h$$
Equality holds when $(2a-2h)=(2a-2h)=h$, when I try to solve, I find:
$$h=frac2a3$$
When I try to do the same with derivatives, I find that the roots of the derivative of $(2a-2h)^2h$ are $a$ and $fraca3$. I may be doing something extremely silly but I can't figure out what is wrong.
derivatives inequality optimization
edited Jul 30 at 9:43
PÃ¥l GD
637413
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
add a comment |Â
up vote
5
down vote
favorite
I have found the following problem:
What is the box (without a top) of largest volume which can be constructed from
a square piece of tin of edge length $2a$ by cutting a square from each corner
and folding up the edges
I have tried to solve it with the AM-GM inequality: I have the sides $(2a-2h), (2a-2h),h$ and then:
$$frac(2a-2h)+ (2a-2h) + h3geq sqrt[3](2a-2h)^2h$$
Equality holds when $(2a-2h)=(2a-2h)=h$, when I try to solve, I find:
$$h=frac2a3$$
When I try to do the same with derivatives, I find that the roots of the derivative of $(2a-2h)^2h$ are $a$ and $fraca3$. I may be doing something extremely silly but I can't figure out what is wrong.
derivatives inequality optimization
edited Jul 30 at 9:43
PÃ¥l GD
637413
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
When you fold with $h=dfrac2a3$, you would know the volume is smaller than you thought.
– Takahiro Waki
Jul 30 at 11:32
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I have found the following problem:
What is the box (without a top) of largest volume which can be constructed from
a square piece of tin of edge length $2a$ by cutting a square from each corner
and folding up the edges
I have tried to solve it with the AM-GM inequality: I have the sides $(2a-2h), (2a-2h),h$ and then:
$$frac(2a-2h)+ (2a-2h) + h3geq sqrt[3](2a-2h)^2h$$
Equality holds when $(2a-2h)=(2a-2h)=h$, when I try to solve, I find:
$$h=frac2a3$$
When I try to do the same with derivatives, I find that the roots of the derivative of $(2a-2h)^2h$ are $a$ and $fraca3$. I may be doing something extremely silly but I can't figure out what is wrong.
derivatives inequality optimization
edited Jul 30 at 9:43
PÃ¥l GD
637413
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
I have found the following problem:
What is the box (without a top) of largest volume which can be constructed from
a square piece of tin of edge length $2a$ by cutting a square from each corner
and folding up the edges
I have tried to solve it with the AM-GM inequality: I have the sides $(2a-2h), (2a-2h),h$ and then:
$$frac(2a-2h)+ (2a-2h) + h3geq sqrt[3](2a-2h)^2h$$
Equality holds when $(2a-2h)=(2a-2h)=h$, when I try to solve, I find:
$$h=frac2a3$$
When I try to do the same with derivatives, I find that the roots of the derivative of $(2a-2h)^2h$ are $a$ and $fraca3$. I may be doing something extremely silly but I can't figure out what is wrong.
derivatives inequality optimization
edited Jul 30 at 9:43
PÃ¥l GD
637413
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
edited Jul 30 at 9:43
PÃ¥l GD
637413
edited Jul 30 at 9:43
PÃ¥l GD
637413
edited Jul 30 at 9:43
PÃ¥l GD
637413
637413
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
asked Jul 30 at 1:54
Billy Rubina
10.1k1254128
10.1k1254128
When you fold with $h=dfrac2a3$, you would know the volume is smaller than you thought.
– Takahiro Waki
Jul 30 at 11:32
add a comment |Â
When you fold with $h=dfrac2a3$, you would know the volume is smaller than you thought.
– Takahiro Waki
Jul 30 at 11:32
When you fold with $h=dfrac2a3$, you would know the volume is smaller than you thought.
– Takahiro Waki
Jul 30 at 11:32
When you fold with $h=dfrac2a3$, you would know the volume is smaller than you thought.
– Takahiro Waki
Jul 30 at 11:32
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
The problem is, that your bound on the left in AM/GM depends on $h$.
Try
$$sqrt[3](2a-2h)^2h=2^1/3sqrt[3](a-h)^22h
le2^1/3frac(a-h)+(a-h)+2h3$$
instead. You get equality when $a-h=2h$ etc.
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
add a comment |Â
up vote
3
down vote
Let $x=$ side of small square $=$ depth of the box;
then $2a-2x=$ side of the square forming bottom of the box, and the volume is $V=(2a-2x)^2x;$ which is the function to be made a maximum by varying $x$.
Applying the rule
$$dfracdVdx=12x^2-16ax+4a^2$$
Solving $12x^2-16ax+4a^2=0$ gives the critical values $x=a,dfrac a3$.
It is evident that $x=a$ must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, $x=dfrac a3$ is found to give a maximum volume of $V=left(2a-2left(dfrac a3right)right)^2dfrac a3=dfrac16a^327$
answered Jul 30 at 2:23
Key Flex
3,955423
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The problem is, that your bound on the left in AM/GM depends on $h$.
Try
$$sqrt[3](2a-2h)^2h=2^1/3sqrt[3](a-h)^22h
le2^1/3frac(a-h)+(a-h)+2h3$$
instead. You get equality when $a-h=2h$ etc.
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
add a comment |Â
up vote
6
down vote
accepted
The problem is, that your bound on the left in AM/GM depends on $h$.
Try
$$sqrt[3](2a-2h)^2h=2^1/3sqrt[3](a-h)^22h
le2^1/3frac(a-h)+(a-h)+2h3$$
instead. You get equality when $a-h=2h$ etc.
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The problem is, that your bound on the left in AM/GM depends on $h$.
Try
$$sqrt[3](2a-2h)^2h=2^1/3sqrt[3](a-h)^22h
le2^1/3frac(a-h)+(a-h)+2h3$$
instead. You get equality when $a-h=2h$ etc.
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
The problem is, that your bound on the left in AM/GM depends on $h$.
Try
$$sqrt[3](2a-2h)^2h=2^1/3sqrt[3](a-h)^22h
le2^1/3frac(a-h)+(a-h)+2h3$$
instead. You get equality when $a-h=2h$ etc.
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 2:01
Lord Shark the Unknown
84.5k950111
84.5k950111
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
add a comment |Â
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
Oh, I guess I understand now: Doing what I did, I just said that the RHS is lesser than $frac4a-3h3$ and not that it is smaller than a single real number.
– Billy Rubina
Jul 30 at 2:47
add a comment |Â
up vote
3
down vote
Let $x=$ side of small square $=$ depth of the box;
then $2a-2x=$ side of the square forming bottom of the box, and the volume is $V=(2a-2x)^2x;$ which is the function to be made a maximum by varying $x$.
Applying the rule
$$dfracdVdx=12x^2-16ax+4a^2$$
Solving $12x^2-16ax+4a^2=0$ gives the critical values $x=a,dfrac a3$.
It is evident that $x=a$ must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, $x=dfrac a3$ is found to give a maximum volume of $V=left(2a-2left(dfrac a3right)right)^2dfrac a3=dfrac16a^327$
answered Jul 30 at 2:23
Key Flex
3,955423
add a comment |Â
up vote
3
down vote
Let $x=$ side of small square $=$ depth of the box;
then $2a-2x=$ side of the square forming bottom of the box, and the volume is $V=(2a-2x)^2x;$ which is the function to be made a maximum by varying $x$.
Applying the rule
$$dfracdVdx=12x^2-16ax+4a^2$$
Solving $12x^2-16ax+4a^2=0$ gives the critical values $x=a,dfrac a3$.
It is evident that $x=a$ must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, $x=dfrac a3$ is found to give a maximum volume of $V=left(2a-2left(dfrac a3right)right)^2dfrac a3=dfrac16a^327$
answered Jul 30 at 2:23
Key Flex
3,955423
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $x=$ side of small square $=$ depth of the box;
then $2a-2x=$ side of the square forming bottom of the box, and the volume is $V=(2a-2x)^2x;$ which is the function to be made a maximum by varying $x$.
Applying the rule
$$dfracdVdx=12x^2-16ax+4a^2$$
Solving $12x^2-16ax+4a^2=0$ gives the critical values $x=a,dfrac a3$.
It is evident that $x=a$ must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, $x=dfrac a3$ is found to give a maximum volume of $V=left(2a-2left(dfrac a3right)right)^2dfrac a3=dfrac16a^327$
answered Jul 30 at 2:23
Key Flex
3,955423
Let $x=$ side of small square $=$ depth of the box;
then $2a-2x=$ side of the square forming bottom of the box, and the volume is $V=(2a-2x)^2x;$ which is the function to be made a maximum by varying $x$.
Applying the rule
$$dfracdVdx=12x^2-16ax+4a^2$$
Solving $12x^2-16ax+4a^2=0$ gives the critical values $x=a,dfrac a3$.
It is evident that $x=a$ must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, $x=dfrac a3$ is found to give a maximum volume of $V=left(2a-2left(dfrac a3right)right)^2dfrac a3=dfrac16a^327$
answered Jul 30 at 2:23
Key Flex
3,955423
edited Jul 30 at 2:29
answered Jul 30 at 2:23
Key Flex
3,955423
answered Jul 30 at 2:23
Key Flex
3,955423
answered Jul 30 at 2:23
Key Flex
3,955423
3,955423
add a comment |Â
add a comment |Â
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When you fold with $h=dfrac2a3$, you would know the volume is smaller than you thought.
– Takahiro Waki
Jul 30 at 11:32