Mixing
Exercise XV num 12 - Calculus Made Easy
$begingroup$
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 Jan 5 at 14:49
LeoBonhartLeoBonhart
134
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:49
LeoBonhartLeoBonhart
134
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:49
LeoBonhartLeoBonhart
134
$endgroup$
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 Jan 5 at 14:49
LeoBonhartLeoBonhart
134
asked Jan 5 at 14:49
LeoBonhartLeoBonhart
134
edited Jan 5 at 15:01
LeoBonhart
asked Jan 5 at 14:49
LeoBonhartLeoBonhart
134
asked Jan 5 at 14:49
LeoBonhartLeoBonhart
134
asked Jan 5 at 14:49
LeoBonhartLeoBonhart
134
134
add a comment |
add a comment |
1 Answer
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$begingroup$
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 Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
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1 Answer
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1 Answer
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$begingroup$
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 Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
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 Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 5 at 15:31
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
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