Question 390537
what would the maximum area of a rectangle with a perimeter of 192km?
<pre>
The rectangle with the most area for a given perimeter is always a 
square.  So each side would be 1/4 of its perimeter making it a 
48km x 48km square whose area is 2304 km². 

But you probably were supposed to show this by using algebra, not just
know that it's a square. So let the two vertical sides be x. Then
the sum of the two horizontal sides would be 192-2x, so each one would
be half that and {{{1/2}}}{192-2x) = 96-x

{{{drawing(400,400,-.5,4.5,-.5,4.5,
locate(4.2,1.5,X),locate(-.2,1.5,X),
locate(1.7,3.3,96-X), locate(1.7,0,96-X)
rectangle(0,0,4,3) )}}}

Let y = the area = length times width = x(96-x)

So the equation for the area y is

     y = x(96 - x)
     y = 96x - x²

Write that as

     y = -x² + 96x 

The graph of that is

{{{graph(400,400,-10,100,-100,2500, 96x-x^2 )  }}}

The area is largest at the vertex (or peak of the graph)

Use the vertex formula:

1. The x-coordinate of the vertex is given by {{{-b/(2a)}}}

2. The y-coordinate is found by substituting the value of the x-coordinate
   into the equation and simplifying.

     y = -x² + 96x is the same as

     y = -x² + 96x + 0  compare to y = ax² + bx + c

     a = -1, b = 96, c = 0

1. The x coordinate of the vertex is 

{{{-b/(2a)=-96/(2*(-1)) = -96/-2 = 48}}}

2. The y-coordinate of the vertex is 

     y = -(48)² + 96(48) 

     y = 2304 km²

Edwin</pre>