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Maths, Addmaths and Statistics: Post your doubts here!

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http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w07_qp_1.pdf
In question 15 , do we have to ensure that the length and the width must equal to 39 since he has a total of 39 fences or we just need to put up some ransom values and then calculate the areas?
Of course we have to keep the total value to 39 fences,but vary their distribution to get the largest area which would be 19 X 10 =190(i dont have super memory about every q i did this one yesterday :p)
 
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But the answer that I got is much larger. :p if the width is 31 and length 8, we get 248, why is this wrong?
If width is 31 then that would mean that the total perimeter of would be 31+31(2 widths)+ 8=70 fences whereas total fences are 39,

You are not focusing clearly,the width is on both sides and we have to take that into consideration.If it is 10 it would be total perimeter of 10+10+19=39 which is largest possible area.Getting me?
 
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If width is 31 then that would mean that the total perimeter of would be 31+31(2 widths)+ 8=70 fences whereas total fences are 39,

You are not focusing clearly,the width is on both sides and we have to take that into consideration.If it is 10 it would be total perimeter of 10+10+19=39 which is largest possible area.Getting me?
Oh yeah, okay, so you mean that the perimeter is giving us the total number of fences, right?
 
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