Exercise XV num 12 - Calculus Made Easy
A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.
Find its dimensions in order that the least amount of iron sheet may be used in its construction.
My approach:
Lets say all dimensions are each $ > 0$;
Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;
Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;
$frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;
$frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;
Is this the right path to continue? Looks too messy in my eyes.
calculus optimization partial-derivative
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LeoBonhartLeoBonhart
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A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.
Find its dimensions in order that the least amount of iron sheet may be used in its construction.
My approach:
Lets say all dimensions are each $ > 0$;
Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;
Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;
$frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;
$frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;
Is this the right path to continue? Looks too messy in my eyes.
calculus optimization partial-derivative
asked 2 days ago
LeoBonhartLeoBonhart
134
New contributor
LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.
Find its dimensions in order that the least amount of iron sheet may be used in its construction.
My approach:
Lets say all dimensions are each $ > 0$;
Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;
Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;
$frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;
$frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;
Is this the right path to continue? Looks too messy in my eyes.
calculus optimization partial-derivative
asked 2 days ago
LeoBonhartLeoBonhart
134
New contributor
LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.
Find its dimensions in order that the least amount of iron sheet may be used in its construction.
My approach:
Lets say all dimensions are each $ > 0$;
Volume of the bucket - $V=frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;
Total exterior area of the bucket - $A= 2frac {hw}{2}+2l(frac{w^2}{4}+h^2)^frac{1}{2} Rightarrow$ $A= hw+frac{4V}{hw}(frac{w^2}{4}+h^2)^frac{1}{2} $;
$frac{partial A}{partial h} = w - frac {4V}{h^2w}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {4V}{w(frac{w^2}{4}+h^2)^frac{1}{2}}$;
$frac{partial A}{partial w} = h - frac {4V}{hw^2}(frac{w^2}{4}+h^2)^frac{1}{2}+frac {V}{h(frac{w^2}{4}+h^2)^frac{1}{2}}$;
Is this the right path to continue? Looks too messy in my eyes.
calculus optimization partial-derivative
calculus optimization partial-derivative
asked 2 days ago
LeoBonhartLeoBonhart
134
New contributor
LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
LeoBonhartLeoBonhart
134
New contributor
LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 days ago
LeoBonhart
asked 2 days ago
LeoBonhartLeoBonhart
134
New contributor
LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
LeoBonhartLeoBonhart
134
asked 2 days ago
LeoBonhartLeoBonhart
134
134
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LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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It is not so messy. Use $w=lambda h$ to get
$$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
$$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
Subtract one from the other to get
$$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
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active
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votes
It is not so messy. Use $w=lambda h$ to get
$$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
$$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
Subtract one from the other to get
$$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
It is not so messy. Use $w=lambda h$ to get
$$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
$$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
Subtract one from the other to get
$$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
It is not so messy. Use $w=lambda h$ to get
$$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
$$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
Subtract one from the other to get
$$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
It is not so messy. Use $w=lambda h$ to get
$$frac{partial A}{partial h} =h lambda -frac{2 lambda V}{h^2 sqrt{lambda ^2+4}}=0implies h -frac{2 V}{h^2 sqrt{lambda ^2+4}}=0tag 1$$
$$frac{partial A}{partial w} =h-frac{8 V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=tag 20$$
Subtract one from the other to get
$$frac{2 left(lambda ^2-4right) V}{h^2 lambda ^2 sqrt{lambda ^2+4}}=0implies lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=lambda h$.
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
answered 2 days ago
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
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