Question 197245
<pre><font size = 4 color = "indigo"><b>
A rectangular field adjacent to the straight bank of a river is to 
be fenced, but there is to be no fencing along the riverbank. If 180 m 
of fencing is available, what is the maximum area that can be 
enclosed?

We draw a rectangle for the area, with a dotted line
for the river bank:

{{{drawing(200,120,2,8,1,5, 
line(3,2,7,2), line(3,2,3,4), line(7,2,7,4),
graph(200,120,2,8,1,5,4*sqrt( sin(6x))/sqrt(sin(6x)) ) )
)}}} 

If we let x ft. be the length of the left and right sides,
the bottom side will be 200 ft - 2x ft.  So we put x on
the two vertical sides and 200-2x on the horizontal side.

{{{drawing(200,120,2,8,1,5, locate(4.5,1.8, "200-2" ),locate(5.7,1.8,x), 
locate(2.7,3.2,x), locate(7.1,3.2,x),
line(3,2,7,2), line(3,2,3,4), line(7,2,7,4),
graph(200,120,2,8,1,5,4*sqrt( sin(6x))/sqrt(sin(6x)) ) )
)}}}

Let y = the area of the rectangle.

Use {{{Area = (base)(height)}}}, so we have

{{{y=(200-2x)(x)}}}

{{{y=x(200-2x)}}}

{{{y=200x-2x^2}}}

{{{y=-2x^2+200x}}}  (Note: this is a quadratic equation)

If we draw the graph of that we get:

{{{graph(400,400,-10,105,-200,5100,200x-2x^2)}}}
   
Since the graph is symmetrical, the highest point
is right in the middle between the two x-intercepts.

So we find the two x-intercepts by setting y=0 and
solving for x:

{{{y=-2x^2+200x}}}
{{{0=-2x^2+200x}}}
{{{2x^2-200x=0}}}

Factor the left side:

{{{2x(x-100)=0}}}

{{{2x=0}}},  {{{x-100=0}}}
{{{x=0}}},   {{{x=100}}}

Halfway between those will be
where x = 50

So the two vertical sides will be 50 feet each,
and the base will be 200 - 2x or

{{{200-2x=200-2(50)=200-100=100}}}

or 100 feet.  And that maximum area will be

(base)(height) = (100)(50) = 5000 square feet.

Edwin</pre>