Answer:
40,000 meters squared
Step-by-step explanation:
Find the optimum and constraint equations. (They are just fancy terms for the two equations you need).
The Constraint is 2x + 2y= 800 (As the plot is a normal rectangle and there is 800 m to make a fence, you would use the perimeter formula for a rectangle)
The optimum is xy= A. The area equals length times width (or in this case) xy.
You will need to plug in the simplified constraint into the optimum.
2y= -2x + 800 (divide by two next)
y= -x + 400
Plug in this to the Optimum
x(-x + 400)
this equals -xsquared + 400x
Find the derivative of it
-2x plus 400
Set it equal to )
x=200, this is only one of the lengths of the side.
In order to find the other, plug in x=200 to the constraint
-x + 400
-200 + 400
y = 200 also
So, going back to the Optimum 200 times 200 equals 40,000.
This means that the maximum area is in fact 40,000 meters squared.
