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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: naved_s9994 on July 05, 2009, 09:45:16 am

Title: MAX min problem!!
Post by: naved_s9994 on July 05, 2009, 09:45:16 am: Mildred wants to construct a rectangular enclosure for her children to play in.

10 M rope

find max area
Title: Re: MAX min problem!!
Post by: kj_ on July 05, 2009, 10:43:49 am: Max area of any rectangular object = square.

Thus, $\frac{10M}{4} = 2.5M$ each side of the square, and thus $2.5 * 2.5 = 6.25m^2$ as max area.

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Of course, you can say $P = x + y$ , $A = xy$ , find out that $A = 5y - y^2$ (because $10 = 2x + 2y$ and thus $x = 5-y$ , then diff A, let it equal 0, and get that y = 2.5 metres. :)
Title: Re: MAX min problem!!
Post by: cobby on July 05, 2009, 10:47:26 am: Quote from: kj_ on July 05, 2009, 10:43:49 am

Of course, you can say P = x + y, A = xy, find out that $A = 5y - y^2$ , then diff A, let it equal 0, and get that y = 2.5 metres. :)
Thats what i did :) :P

$2x + 2y = 10$
$x + y = 5$
$\implies y = 5 - x$
$A = x*y$
$(5-x) * x$
$5x- x^2$

Sketch and find max $(2.5,6.25)$
Title: Re: MAX min problem!!
Post by: naved_s9994 on July 05, 2009, 11:07:15 am: yea guys i got same answer as you..lol thnx!

but i was wondering cant we let derivate = 0
Title: Re: MAX min problem!!
Post by: cobby on July 05, 2009, 11:40:03 am: Quote from: naved_s9994 on July 05, 2009, 11:07:15 am
yea guys i got same answer as you..lol thnx!

but i was wondering cant we let derivate = 0

yeh :)
Title: Re: MAX min problem!!
Post by: kj_ on July 05, 2009, 12:03:35 pm: Quote from: naved_s9994 on July 05, 2009, 11:07:15 am
yea guys i got same answer as you..lol thnx!

but i was wondering cant we let derivate = 0

Yes, it's exactly what I did in my 2nd step, read it again :)
Title: Re: MAX min problem!!
Post by: naved_s9994 on July 08, 2009, 03:39:01 pm: ahh thnx kj_ !