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C2 Calculating Area...
This is a differentiation question, but I can't get the first part >< .
Diagram: http://i43.tinypic.com/t5lqhf.jpg
The shape in the diagram above is a wire frame in the form of a large rectangle split by parallel lengths of wire into 12 smaller equal sized rectangles. Given that the total length of wire is 1512mm, show that the area of the whole shape is A mm2 where where x mm is the width of one of the smaller rectangles. Find the maximum area which can be enclosed in this way.
My method:
My answer is double the actual answer >< , what did I do wrong?
Diagram: http://i43.tinypic.com/t5lqhf.jpg
The shape in the diagram above is a wire frame in the form of a large rectangle split by parallel lengths of wire into 12 smaller equal sized rectangles. Given that the total length of wire is 1512mm, show that the area of the whole shape is A mm2 where where x mm is the width of one of the smaller rectangles. Find the maximum area which can be enclosed in this way.
My method:
Unparseable latex formula:
\begin{matrix}[br]18x+7y=1512\\[br]A=2y\times 6x=12xy\\[br]y=216-\frac{18}{7}x\\[br]12x(216-\frac{18}{7})\\=2592x-\frac{216}{7}x^2[br]\end{matrix}
My answer is double the actual answer >< , what did I do wrong?
Reply 1
Ok, I've tried numerous times and keep getting the same result.
Reply 2
its 14y
Reply 3
It's amazing when you see a thread in the sidebar, think "I can help with that!" then realise it's your own one from those couple years ago xD.
Original post by ViralRiver
It's amazing when you see a thread in the sidebar, think "I can help with that!" then realise it's your own one from those couple years ago xD.
why not actually write help for others in future instead of this?
Original post by marcy
its 14y
Why though?
Reply 6
Original post by ithinkso
Why though?
There are 7 bars of length y along the top and 7 along the bottom. If that makes sense.
Original post by Bjornhattan
There are 7 bars of length y along the top and 7 along the bottom. If that makes sense.
no because y is the height and they're not stated to be square in the problem
Reply 8
Original post by ithinkso
no because y is the height and they're not stated to be square in the problem
They are of equal size, however, so you can assume that all vertical lines are of length y. There are 14 vertical lines.
Original post by Bjornhattan
They are of equal size, however, so you can assume that all vertical lines are of length y. There are 14 vertical lines.
Right I thought that this was the question http://i.imgur.com/E3ylLpb.png
which doesn't state they're square
Reply 10
If all of the rectangles are of equal size though, all of their dimensions are x by y.
Original post by Bjornhattan
If all of the rectangles are of equal size though, all of their dimensions are x by y.
sorry even size but they're not square... anyway my point was that noone actually provided a decent explanation for the problem so that others (ie me) who find it in future are none the wiser
Reply 12
Original post by ithinkso
sorry even size but they're not square... anyway my point was that noone actually provided a decent explanation for the problem so that others (ie me) who find it in future are none the wiser
Sorry, don't worry about it.
Original post by Bjornhattan
Sorry, don't worry about it.
I'm just posting to prompt the OPs to actually amend the post so it has some value seeing as it comes up in a search for this question
Original post by ithinkso
I'm just posting to prompt the OPs to actually amend the post so it has some value seeing as it comes up in a search for this question
do you know how long ago this thread was made?
Original post by TeeEm
do you know how long ago this thread was made?
didn't consider that, just found it in a search and left a comment
Original post by ithinkso
didn't consider that, just found it in a search and left a comment
it happens sometimes.
Original post by TeeEm
it happens sometimes.
TBH I would have probably posted anyway - as this problem came up on google and I felt that there should be some better information here than there was.
I don't mean this to 'necro' or whatever though, more in a helpful sense than anything. If this etiquette is generally frowned on here I'll bear it in mind for future
Original post by ViralRiver
This is a differentiation question, but I can't get the first part >< .
Diagram: http://i43.tinypic.com/t5lqhf.jpg
The shape in the diagram above is a wire frame in the form of a large rectangle split by parallel lengths of wire into 12 smaller equal sized rectangles. Given that the total length of wire is 1512mm, show that the area of the whole shape is A mm2 where where x mm is the width of one of the smaller rectangles. Find the maximum area which can be enclosed in this way.
My method:
My answer is double the actual answer >< , what did I do wrong?
Diagram: http://i43.tinypic.com/t5lqhf.jpg
The shape in the diagram above is a wire frame in the form of a large rectangle split by parallel lengths of wire into 12 smaller equal sized rectangles. Given that the total length of wire is 1512mm, show that the area of the whole shape is A mm2 where where x mm is the width of one of the smaller rectangles. Find the maximum area which can be enclosed in this way.
My method:
Unparseable latex formula:
\begin{matrix}[br]18x+7y=1512\\[br]A=2y\times 6x=12xy\\[br]y=216-\frac{18}{7}x\\[br]12x(216-\frac{18}{7})\\=2592x-\frac{216}{7}x^2[br]\end{matrix}
My answer is double the actual answer >< , what did I do wrong?
Ok not sure if anyone is interested but this question just came up in an assignment in 2020 so its still clearly doing bits in a level.
heres my solution
1512=14+18x
1512-18x=14y
108-9/7x=y
A=2y x 6x
A=12xy
=12(x)(108-9/7x)
=12(108x-9/7x^2)
=1296x-108/7x^2
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