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.

there are 25 people coming to the party

step-by-step explanation:

the text says that there will be 22 children at this party, 1 teacher, and 2 parents. so if you add them all together (22+1+2) it will equal 25. (=25)