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Calculus - Maximum volume equation
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I tried my best to come up with an answer but I failed. I would really appreciate if someone can help me with this:
Question: To carry a suitcase on an airplace, the length + width + height must be less than or equal to $ 62$ in. Assuming that height is fixed, show that the maximum volume is $ V = h(31-frac12h)^2 .$
Based on that, I did the following:
$$l+w+h = 62$$
$$l+w = 62 - h$$
$$V = lwh$$
$$dV = dl(lwh) + dw(lwh) +dh(lwh)$$
$$dV = wh + lh + 0$$
$$dV = h(l+w)$$
$$dV = h(62-h)$$
I dont know what else to do after this...Any suggestions recommendation is greatly appreciated!
calculus
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
asked Jul 21 at 6:30
boniface316
1176
add a comment |Â
up vote
0
down vote
favorite
I tried my best to come up with an answer but I failed. I would really appreciate if someone can help me with this:
Question: To carry a suitcase on an airplace, the length + width + height must be less than or equal to $ 62$ in. Assuming that height is fixed, show that the maximum volume is $ V = h(31-frac12h)^2 .$
Based on that, I did the following:
$$l+w+h = 62$$
$$l+w = 62 - h$$
$$V = lwh$$
$$dV = dl(lwh) + dw(lwh) +dh(lwh)$$
$$dV = wh + lh + 0$$
$$dV = h(l+w)$$
$$dV = h(62-h)$$
I dont know what else to do after this...Any suggestions recommendation is greatly appreciated!
calculus
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
asked Jul 21 at 6:30
boniface316
1176
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 21 at 6:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I tried my best to come up with an answer but I failed. I would really appreciate if someone can help me with this:
Question: To carry a suitcase on an airplace, the length + width + height must be less than or equal to $ 62$ in. Assuming that height is fixed, show that the maximum volume is $ V = h(31-frac12h)^2 .$
Based on that, I did the following:
$$l+w+h = 62$$
$$l+w = 62 - h$$
$$V = lwh$$
$$dV = dl(lwh) + dw(lwh) +dh(lwh)$$
$$dV = wh + lh + 0$$
$$dV = h(l+w)$$
$$dV = h(62-h)$$
I dont know what else to do after this...Any suggestions recommendation is greatly appreciated!
calculus
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
asked Jul 21 at 6:30
boniface316
1176
I tried my best to come up with an answer but I failed. I would really appreciate if someone can help me with this:
Question: To carry a suitcase on an airplace, the length + width + height must be less than or equal to $ 62$ in. Assuming that height is fixed, show that the maximum volume is $ V = h(31-frac12h)^2 .$
Based on that, I did the following:
$$l+w+h = 62$$
$$l+w = 62 - h$$
$$V = lwh$$
$$dV = dl(lwh) + dw(lwh) +dh(lwh)$$
$$dV = wh + lh + 0$$
$$dV = h(l+w)$$
$$dV = h(62-h)$$
I dont know what else to do after this...Any suggestions recommendation is greatly appreciated!
calculus
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
asked Jul 21 at 6:30
boniface316
1176
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
edited Jul 21 at 7:08
Gerry Myerson
143k7144294
143k7144294
asked Jul 21 at 6:30
boniface316
1176
asked Jul 21 at 6:30
boniface316
1176
asked Jul 21 at 6:30
boniface316
1176
1176
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 21 at 6:32
add a comment |Â
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 21 at 6:32
1
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 21 at 6:32
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 21 at 6:32
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
We want to express the volume as a function of a single variable. Since $h$ is a constant, that variable can be either $l$ or $w$. Let's use $w$.
We are given that $l + w + h = 62~textin$. Hence, $l = 62~textin - h - w$. Substituting $62~textin - h - w$ for $l$ in the equation $V = lwh$ yields
$$V(w) = (62~textin - h - w)wh = (62~textin)wh - wh^2 - w^2h$$
Differentiating with respect to $w$, while remembering that $h$ is a constant, yields
$$V'(w) = (62~textin)h - h^2 - 2wh$$
Setting the derivative equal to $0$ and solving for $w$ yields
$$w = 31~textin - frach2$$
Since $h > 0$,
$$V''(w) = -2h < 0$$
Thus, the function has a relative maximum at $w = 31~textin - dfrach2$ by the Second Derivative Test.
When $w = 31~textin - dfrach2$, then
$$l = 62~textin - h - w = 62~textin - h - left(31~textin - frach2right) = 31~textin - frach2$$
Thus, the maximum volume of the suitcase is
$$V = lwh = left(31~textin - frach2right)left(31~textin - frach2right)h = left(31~textin - frach2right)^2h$$
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
add a comment |Â
up vote
1
down vote
$$l+w+h=62$$
Therefore, $L+w=62-h$. In order to maximise the volume, the area enclosed by the rectangle, $ltimes w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-frac 1 2 h)^2h$.
QED.
answered Jul 21 at 8:53
Dr Peter McGowan
4637
add a comment |Â
up vote
1
down vote
Using Cauchy-Schwarz inequality
$ l+w+h = 62$
$l+w = 62-h$
$(l^2+w^2)(1^2+1^2) geq (l+w)^2$
$2(l^2+w^2) geq (l+w)^2$
$2((l+w)^2-2lw) geq (l+w)^2$
$2(l+w)^2-4lw geq (l+w)^2$
$(l+w)^2 geq 4lw$
since $ lw = fracVh $
$(62-h)^2 geq frac4Vh$
$h(62-h)^2 geq 4V$
$ V leq h(31-frac12h)^2$
answered Jul 21 at 8:54
Pizzaroot
1056
add a comment |Â
up vote
0
down vote
Hint
To maximise the volume at fixed height you have to maximise the section of the suitcase. In this case it has to be either rectangular of square: in any case you can construct a square with the same area as a rectangle. Square section is very much easier to work with.
Solution
To maximise the volume with a fixed height you have to maximise the section of the suitcase. In the simplest case, because it doesn't really matters, you can choose a square section so that $w=l$. In this case you have $$2w = 62-h\w = 31-1over 2h$$ the area of a square is $$S = w^2 = (31-1over 2h)^2$$ In the end, the maximum volume is $$V=hS = h(31-1over 2h)^2$$
answered Jul 21 at 8:25
Davide Morgante
1,812220
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We want to express the volume as a function of a single variable. Since $h$ is a constant, that variable can be either $l$ or $w$. Let's use $w$.
We are given that $l + w + h = 62~textin$. Hence, $l = 62~textin - h - w$. Substituting $62~textin - h - w$ for $l$ in the equation $V = lwh$ yields
$$V(w) = (62~textin - h - w)wh = (62~textin)wh - wh^2 - w^2h$$
Differentiating with respect to $w$, while remembering that $h$ is a constant, yields
$$V'(w) = (62~textin)h - h^2 - 2wh$$
Setting the derivative equal to $0$ and solving for $w$ yields
$$w = 31~textin - frach2$$
Since $h > 0$,
$$V''(w) = -2h < 0$$
Thus, the function has a relative maximum at $w = 31~textin - dfrach2$ by the Second Derivative Test.
When $w = 31~textin - dfrach2$, then
$$l = 62~textin - h - w = 62~textin - h - left(31~textin - frach2right) = 31~textin - frach2$$
Thus, the maximum volume of the suitcase is
$$V = lwh = left(31~textin - frach2right)left(31~textin - frach2right)h = left(31~textin - frach2right)^2h$$
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
add a comment |Â
up vote
1
down vote
accepted
We want to express the volume as a function of a single variable. Since $h$ is a constant, that variable can be either $l$ or $w$. Let's use $w$.
We are given that $l + w + h = 62~textin$. Hence, $l = 62~textin - h - w$. Substituting $62~textin - h - w$ for $l$ in the equation $V = lwh$ yields
$$V(w) = (62~textin - h - w)wh = (62~textin)wh - wh^2 - w^2h$$
Differentiating with respect to $w$, while remembering that $h$ is a constant, yields
$$V'(w) = (62~textin)h - h^2 - 2wh$$
Setting the derivative equal to $0$ and solving for $w$ yields
$$w = 31~textin - frach2$$
Since $h > 0$,
$$V''(w) = -2h < 0$$
Thus, the function has a relative maximum at $w = 31~textin - dfrach2$ by the Second Derivative Test.
When $w = 31~textin - dfrach2$, then
$$l = 62~textin - h - w = 62~textin - h - left(31~textin - frach2right) = 31~textin - frach2$$
Thus, the maximum volume of the suitcase is
$$V = lwh = left(31~textin - frach2right)left(31~textin - frach2right)h = left(31~textin - frach2right)^2h$$
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We want to express the volume as a function of a single variable. Since $h$ is a constant, that variable can be either $l$ or $w$. Let's use $w$.
We are given that $l + w + h = 62~textin$. Hence, $l = 62~textin - h - w$. Substituting $62~textin - h - w$ for $l$ in the equation $V = lwh$ yields
$$V(w) = (62~textin - h - w)wh = (62~textin)wh - wh^2 - w^2h$$
Differentiating with respect to $w$, while remembering that $h$ is a constant, yields
$$V'(w) = (62~textin)h - h^2 - 2wh$$
Setting the derivative equal to $0$ and solving for $w$ yields
$$w = 31~textin - frach2$$
Since $h > 0$,
$$V''(w) = -2h < 0$$
Thus, the function has a relative maximum at $w = 31~textin - dfrach2$ by the Second Derivative Test.
When $w = 31~textin - dfrach2$, then
$$l = 62~textin - h - w = 62~textin - h - left(31~textin - frach2right) = 31~textin - frach2$$
Thus, the maximum volume of the suitcase is
$$V = lwh = left(31~textin - frach2right)left(31~textin - frach2right)h = left(31~textin - frach2right)^2h$$
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
We want to express the volume as a function of a single variable. Since $h$ is a constant, that variable can be either $l$ or $w$. Let's use $w$.
We are given that $l + w + h = 62~textin$. Hence, $l = 62~textin - h - w$. Substituting $62~textin - h - w$ for $l$ in the equation $V = lwh$ yields
$$V(w) = (62~textin - h - w)wh = (62~textin)wh - wh^2 - w^2h$$
Differentiating with respect to $w$, while remembering that $h$ is a constant, yields
$$V'(w) = (62~textin)h - h^2 - 2wh$$
Setting the derivative equal to $0$ and solving for $w$ yields
$$w = 31~textin - frach2$$
Since $h > 0$,
$$V''(w) = -2h < 0$$
Thus, the function has a relative maximum at $w = 31~textin - dfrach2$ by the Second Derivative Test.
When $w = 31~textin - dfrach2$, then
$$l = 62~textin - h - w = 62~textin - h - left(31~textin - frach2right) = 31~textin - frach2$$
Thus, the maximum volume of the suitcase is
$$V = lwh = left(31~textin - frach2right)left(31~textin - frach2right)h = left(31~textin - frach2right)^2h$$
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
answered Jul 21 at 9:02
N. F. Taussig
38.2k93053
38.2k93053
add a comment |Â
add a comment |Â
up vote
1
down vote
$$l+w+h=62$$
Therefore, $L+w=62-h$. In order to maximise the volume, the area enclosed by the rectangle, $ltimes w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-frac 1 2 h)^2h$.
QED.
answered Jul 21 at 8:53
Dr Peter McGowan
4637
add a comment |Â
up vote
1
down vote
$$l+w+h=62$$
Therefore, $L+w=62-h$. In order to maximise the volume, the area enclosed by the rectangle, $ltimes w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-frac 1 2 h)^2h$.
QED.
answered Jul 21 at 8:53
Dr Peter McGowan
4637
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$l+w+h=62$$
Therefore, $L+w=62-h$. In order to maximise the volume, the area enclosed by the rectangle, $ltimes w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-frac 1 2 h)^2h$.
QED.
answered Jul 21 at 8:53
Dr Peter McGowan
4637
$$l+w+h=62$$
Therefore, $L+w=62-h$. In order to maximise the volume, the area enclosed by the rectangle, $ltimes w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-frac 1 2 h)^2h$.
QED.
answered Jul 21 at 8:53
Dr Peter McGowan
4637
answered Jul 21 at 8:53
Dr Peter McGowan
4637
answered Jul 21 at 8:53
Dr Peter McGowan
4637
answered Jul 21 at 8:53
Dr Peter McGowan
4637
4637
add a comment |Â
add a comment |Â
up vote
1
down vote
Using Cauchy-Schwarz inequality
$ l+w+h = 62$
$l+w = 62-h$
$(l^2+w^2)(1^2+1^2) geq (l+w)^2$
$2(l^2+w^2) geq (l+w)^2$
$2((l+w)^2-2lw) geq (l+w)^2$
$2(l+w)^2-4lw geq (l+w)^2$
$(l+w)^2 geq 4lw$
since $ lw = fracVh $
$(62-h)^2 geq frac4Vh$
$h(62-h)^2 geq 4V$
$ V leq h(31-frac12h)^2$
answered Jul 21 at 8:54
Pizzaroot
1056
add a comment |Â
up vote
1
down vote
Using Cauchy-Schwarz inequality
$ l+w+h = 62$
$l+w = 62-h$
$(l^2+w^2)(1^2+1^2) geq (l+w)^2$
$2(l^2+w^2) geq (l+w)^2$
$2((l+w)^2-2lw) geq (l+w)^2$
$2(l+w)^2-4lw geq (l+w)^2$
$(l+w)^2 geq 4lw$
since $ lw = fracVh $
$(62-h)^2 geq frac4Vh$
$h(62-h)^2 geq 4V$
$ V leq h(31-frac12h)^2$
answered Jul 21 at 8:54
Pizzaroot
1056
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using Cauchy-Schwarz inequality
$ l+w+h = 62$
$l+w = 62-h$
$(l^2+w^2)(1^2+1^2) geq (l+w)^2$
$2(l^2+w^2) geq (l+w)^2$
$2((l+w)^2-2lw) geq (l+w)^2$
$2(l+w)^2-4lw geq (l+w)^2$
$(l+w)^2 geq 4lw$
since $ lw = fracVh $
$(62-h)^2 geq frac4Vh$
$h(62-h)^2 geq 4V$
$ V leq h(31-frac12h)^2$
answered Jul 21 at 8:54
Pizzaroot
1056
Using Cauchy-Schwarz inequality
$ l+w+h = 62$
$l+w = 62-h$
$(l^2+w^2)(1^2+1^2) geq (l+w)^2$
$2(l^2+w^2) geq (l+w)^2$
$2((l+w)^2-2lw) geq (l+w)^2$
$2(l+w)^2-4lw geq (l+w)^2$
$(l+w)^2 geq 4lw$
since $ lw = fracVh $
$(62-h)^2 geq frac4Vh$
$h(62-h)^2 geq 4V$
$ V leq h(31-frac12h)^2$
answered Jul 21 at 8:54
Pizzaroot
1056
answered Jul 21 at 8:54
Pizzaroot
1056
answered Jul 21 at 8:54
Pizzaroot
1056
answered Jul 21 at 8:54
Pizzaroot
1056
1056
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint
To maximise the volume at fixed height you have to maximise the section of the suitcase. In this case it has to be either rectangular of square: in any case you can construct a square with the same area as a rectangle. Square section is very much easier to work with.
Solution
To maximise the volume with a fixed height you have to maximise the section of the suitcase. In the simplest case, because it doesn't really matters, you can choose a square section so that $w=l$. In this case you have $$2w = 62-h\w = 31-1over 2h$$ the area of a square is $$S = w^2 = (31-1over 2h)^2$$ In the end, the maximum volume is $$V=hS = h(31-1over 2h)^2$$
answered Jul 21 at 8:25
Davide Morgante
1,812220
add a comment |Â
up vote
0
down vote
Hint
To maximise the volume at fixed height you have to maximise the section of the suitcase. In this case it has to be either rectangular of square: in any case you can construct a square with the same area as a rectangle. Square section is very much easier to work with.
Solution
To maximise the volume with a fixed height you have to maximise the section of the suitcase. In the simplest case, because it doesn't really matters, you can choose a square section so that $w=l$. In this case you have $$2w = 62-h\w = 31-1over 2h$$ the area of a square is $$S = w^2 = (31-1over 2h)^2$$ In the end, the maximum volume is $$V=hS = h(31-1over 2h)^2$$
answered Jul 21 at 8:25
Davide Morgante
1,812220
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint
To maximise the volume at fixed height you have to maximise the section of the suitcase. In this case it has to be either rectangular of square: in any case you can construct a square with the same area as a rectangle. Square section is very much easier to work with.
Solution
To maximise the volume with a fixed height you have to maximise the section of the suitcase. In the simplest case, because it doesn't really matters, you can choose a square section so that $w=l$. In this case you have $$2w = 62-h\w = 31-1over 2h$$ the area of a square is $$S = w^2 = (31-1over 2h)^2$$ In the end, the maximum volume is $$V=hS = h(31-1over 2h)^2$$
answered Jul 21 at 8:25
Davide Morgante
1,812220
Hint
To maximise the volume at fixed height you have to maximise the section of the suitcase. In this case it has to be either rectangular of square: in any case you can construct a square with the same area as a rectangle. Square section is very much easier to work with.
Solution
To maximise the volume with a fixed height you have to maximise the section of the suitcase. In the simplest case, because it doesn't really matters, you can choose a square section so that $w=l$. In this case you have $$2w = 62-h\w = 31-1over 2h$$ the area of a square is $$S = w^2 = (31-1over 2h)^2$$ In the end, the maximum volume is $$V=hS = h(31-1over 2h)^2$$
answered Jul 21 at 8:25
Davide Morgante
1,812220
answered Jul 21 at 8:25
Davide Morgante
1,812220
answered Jul 21 at 8:25
Davide Morgante
1,812220
answered Jul 21 at 8:25
Davide Morgante
1,812220
1,812220
add a comment |Â
add a comment |Â
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1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 21 at 6:32