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This Is Not An Exam Question And I Think I Know The Answer!!
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A rope 120 metres long can be formed into any shape so long as the 2 ends are touching. Find the maximum area that can be enclosed by the shape.
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And that's why you find catering size cans of veg at the foot of beanstalks.
22:54 Thu 20th Apr 2023
The area will be the same regardless of the shape. so take the easiest shape to work with as being a circle with a circumference of 120M, from this apply the formula for the circumference of a circle which will give you the radius/diameter. Then apply that to the formula for the area of a circle.
You should either know or be able to find out the formula for the area/circumference of a circle.
You should either know or be able to find out the formula for the area/circumference of a circle.
"The area will be the same regardless of the shape."
This isn't remotely true. Take a rectangular shape of sides 59.9 metres by 0.1m (area 5.9m^2) and compare to a square shape of sides 30m (area 900m^2).
This is a famous optimisation problem -- for a fixed perimeter, the maximum area that can be bounded is a circle.
This isn't remotely true. Take a rectangular shape of sides 59.9 metres by 0.1m (area 5.9m^2) and compare to a square shape of sides 30m (area 900m^2).
This is a famous optimisation problem -- for a fixed perimeter, the maximum area that can be bounded is a circle.
If the only reason for cylindrical food tins is to minimise the amount of steel used, then the aspect ratio (diameter to length) would similarly be optimised – but this is not the case.
The optimum aspect ratio results in the type of tin that is seen containing fish (somewhat squat/flat); if this was to be applied to larger tins the resultant diameter would be difficult to hold in one hand.
So most food tins have a diameter determined by ergonomics related to humans handling the tin, rather than the minimum amount of metal required to make the tin.
The optimum aspect ratio results in the type of tin that is seen containing fish (somewhat squat/flat); if this was to be applied to larger tins the resultant diameter would be difficult to hold in one hand.
So most food tins have a diameter determined by ergonomics related to humans handling the tin, rather than the minimum amount of metal required to make the tin.
yeah solve 2-pi-arr =120
Proof? I bet Clare has a better one
1965 -WE were told the proof was thus: (*)
It has to be symmetrical ( not sure why) and so square wd be OK but not rectangle. - ( actually you can jump to circle by that alone)
and it is a polygon, with no re-entrant bit zit, U shape- if there is, short cirucit and - draw line across and enlarge area thereby
and the polygons can be made smaller side and larger area - and in the end you get a circle.
we all went ugggh
this is a topological proof which obviously doesnt appeal to 14 y olds
(*) Proof what den? Clare hasnt met the whizzy one liner AB put-down yet. Proof that the largest area is a cirlce
other than the proof: well it just is.
Proof? I bet Clare has a better one
1965 -WE were told the proof was thus: (*)
It has to be symmetrical ( not sure why) and so square wd be OK but not rectangle. - ( actually you can jump to circle by that alone)
and it is a polygon, with no re-entrant bit zit, U shape- if there is, short cirucit and - draw line across and enlarge area thereby
and the polygons can be made smaller side and larger area - and in the end you get a circle.
we all went ugggh
this is a topological proof which obviously doesnt appeal to 14 y olds
(*) Proof what den? Clare hasnt met the whizzy one liner AB put-down yet. Proof that the largest area is a cirlce
other than the proof: well it just is.
I'm not proving this. Not on AB, at least, the syntax I'd need isn't supported. Also, as posed, I think you only need to *know* what the max area shape should be in this situation, rather than set up the complete problem and consider all possible shapes, which I think goes far beyond the realms of A level techniques.
A cylinder with the least area.....
No - or yes - that is the gold-plated baked bean can problem.
1965 again. - gold plate to minimise the use of gold
can be reduced to one independent variable - suitable for school calculus
otherwise partial differentials are used ( V as a min for both r and h).
The maff master told us the element of a volume of revolution was pi-y2- delta x on the last day of one term
and asked us on the first day of the next.... I recalled pi y2 dee-x
and he went christ. Notice school boy error of delta x confused with dee-x - - oh dear dear ! I got addl-maff O level when I was 14
No - or yes - that is the gold-plated baked bean can problem.
1965 again. - gold plate to minimise the use of gold
can be reduced to one independent variable - suitable for school calculus
otherwise partial differentials are used ( V as a min for both r and h).
The maff master told us the element of a volume of revolution was pi-y2- delta x on the last day of one term
and asked us on the first day of the next.... I recalled pi y2 dee-x
and he went christ. Notice school boy error of delta x confused with dee-x - - oh dear dear ! I got addl-maff O level when I was 14
// contain the maximum volume with a minimum amount of material would use a sphere – imagine that on supermarket shelves //
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