What should maximum dimensions be to enclose the maximum area using all available fencing?

You have 640m of fencing available. You need to fence in a rectangular area of fields that is then subdivided into three parts with fencing parallel to one of the sides of the rectangle. What should the dimensions be to enclose the maximum area using all of the available fencing?|||Ok...this should be good. Here is a diagram:


http://farm4.static.flickr.com/3269/3098...


Here:


x = length


y = width (also the side parallel to where it is all divided)





Given:


Amount of fence = 640


Need max area.





In this example the total amount of fence used is:


fence_used = perimeter of field + fence used inside enclosure


640 = (2*x + 2*y) + ( 2y )


640 = 2x + 4y


640 - 2x = 4 y


y = 160 - (1/2)*x





Area becomes:


area = x*y


area(x) = x*(160 - (1/2)*x)


area(x) = 160x -(1/2)*x^2


To find max and min of a function find area%26#039;(x), make it equal to zero and solve for x:


area%26#039;(x) = 160 - x


0 = 160 - x


x = 160 --%26gt; this should be your max value





Now since:


y = 160 - (1/2)*x


Plug in x:


y = 160-80 = 80





For max area the dimensions should be:


x = 160


y = 80





Check:


640 = 2x + 4y


2*(160) + 4*(80) = 640


LS == RS





Correct.|||Let%26#039;s look at what you%26#039;ve got. You%26#039;ve got 640 m of fencing, and you have to use all of it. Let%26#039;s draw a picture.





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Kind of, except without the dots in between.





You have two long sides. And, you have 4 short sides. Normally, to find the perimeter of a rectangle, you simply do 2L + 2W. But, we don%26#039;t have just 2 widths... we have 4 widths.





So...





2L + 4W = 640





But, we can%26#039;t solve that because it has two variables. We need another equation.





If you look at your picture, perhaps two of the widths add up to one of the lengths... yeah? Let%26#039;s try that and see if it works.





2W = L





Substitute 2W for the L in the first equation.





2(2W) + 4W = 640


4W + 4W = 640


8W = 640


W = 80





If W = 80, then...





2W = L


2(80) = L


160 = L





Let%26#039;s test and see if these dimensions fulfill the requirements. We have four widths, and if each one is 80 m, then we have used 320 m in fencing with the widths. If we have two lengths, nad each one is 160 m, then we have used 320 m in fencing there.





320 m + 320 m = 640 m





It works. So, the dimensions are 160 m x 80 m.|||Total length of fence (P) = (perimeter of bounding box) + (length of first subdividing fence) + (length of second subdividing fence)


= 2*L + 2*W + W + W


= 2*L + 4*W





i.e. L = (P - 4W) / 2





Constraint: P = 640m





Area = L * W


= W * (P - 4W) / 2





Area extremum when dA/dW = 0





dA = 0.5 [(P - 4W) + W*(-4)] dW = 0.5 [P - 8W] dW





dA = 0 : W = P / 8


so


L = (P - 4W) / 2 = 0.5 P (1 - 0.5) = (1/4) P|||its_victoria08, how do you know that 2W = L. You drew the picture and just assumed that without any valid reasons.