Answer to Question #341247 in Calculus for Stella

Question

A rectangular field is to be enclosed and divided into four equal lots by fences parallel to one of the sides. a total of 10,000 meters of fence are available. Find the area of the largest field that can be enclosed.

Answer 1

The total length will be x

and the height will be y

Needed Equations:

Perimeter of this diagram

10000=2x+5y

Total Area A=xy

Solve for y using the equation

10000=2x+5y

5y=10000-2x

y=2000-2/5x

Substitute the equation for y into the function for area.

A=x(2000-2/5x)=2000x-2/5x²

Find the derivative of the equation for area.

A'=2000-4/5x

Use the derivative equation in order to find the critical point(s) that maximize the area.

Critical points are when

A'=0 and when A' does not exist. It is also good to check the endpoints of an equation in order to check for a maximum or minimum.Since A' always exists, only find where A'=0

(there will be no endpoints to check since this is a field).

0=2000 -4/5x

x=2000/(4/5)=2500

A'is positive when x<2500 and A' is negative when x>2500 , therefore meaning that x=2500 is a maximum. Since this value is a maximum, the area is maximized when the total length is 2500 m.

Find the height ( y ) when x=2500

y=2000-2/5x=2000-2/5(2500)=1000 m

The dimensions that will maximize the area the total area of the pig pen will be

2500m by 1000 m

A=2500x1000=2.5x10⁵ m2