Afarmer wishes to build a rectangular plot with 800 meters of wire fencing. use calculus to find the maximum area that can be enclosed by the fence

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.

8496 cars

step-by-step explanation:

you can't solve x because it is 2 linear equation and you only have one equation