Question 353187
A rancher uses 834 meters of fencing to enclose a rectangular region and also to subdivide the region into two smaller rectangular regions by placing a fence parallel to one of the sides. Find the dimension that produce the greastest enclosed area.

<pre>

We might as well put the subdivision right in the middle and 
divide the rectangle into two equal x by h rectangles.


{{{drawing(400,275,-16,16,-11,11, rectangle(-15,-10,0,10),
 rectangle(0,-10,15,10), locate(-15+.2,0,h), locate(0+.2,0,h), locate(15+.2,0,h), locate(7.5,10,x), locate(-7.5,10,x), locate(7.5,-10,x), locate(-7.5,-10,x)


  )}}}
 
Let the area be {{{y}}}

Area = (base)(height)

Base = {{{2x}}}
Height = {{{h}}}

Let the area be {{{y}}}

{{{y = 2xh}}}

Sum of fencings = {{{4x + 3h}}}

{{{4x + 3h = 834}}}
     
Solve for h

{{{3h = 834 - 4x}}}

{{{h = (834-4x)/3}}}

{{{h = 843/3 - expr(4/3)x}}}

{{{h = 281 - expr(4/3)x}}}

Substitute in

{{{y = 2xh}}}

{{{y = 2x(281 - expr(4/3)x)}}}

{{{y = 562x - expr(8/3)x^2)}}}

{{{y = 562x - expr(8/3)x^2)}}}

{{{y = -expr(8/3)x^2 + 562x}}}

Use the vertex formula for this parabola:

{{{graph(400,400,-10,220,-1000,31000, 562x -(8/3)*x^2)}}}

x-coordinate of vertex = {{{-b/(2a)=(-562)/(2*(-8/3))=(-562)/(-16/3)=-562*(-3/16)=843/8=105&3/8}}}

{{{h = (834-4x)/3}}}
{{{h = (834-4*expr(843/8))/3}}}
{{{h = (834-843/2)/3}}}
{{{h = (834-843/2)*(1/3)}}}
{{{h = 281-843/6}}}
{{{h = 1686/6 -843/6}}}
{{{h = 843/6}}}
{{{h = 140&1/2}}}

So the dimensions are 2x by h or {{{2(843/8)}}} by {{{843/6}}} or 

{{{843/4}}}m by {{{843/6}}}m

or

210.75m by 140.5m


{{{drawing(400,275,-16,16,-11,11, rectangle(-15,-10,0,10),
 rectangle(0,-10,15,10), locate(-15+.2,0,140.5), locate(0+.2,0,140.5), locate(11.5,0,140.5), locate(7.5,10,105.375), locate(-7.5,10,105.375), locate(7.5,-10,105.375), locate(-7.5,-10,105.375)


  )}}}
 

Edwin</pre>