Answer to Question #337520 in Calculus for dme

Question

Albert builds rectangular pig pen with two parallel partitions using 300 meters of

fencing materials. What should be the dimensions of the pig pen so that it would enclose the largest

possible area?

Answer 1

The total length will be x

and the height will be y

Needed Equations:

Perimeter of this diagram

300=3x+4y

Total Area A=xy

Solve for y using the equation

300=3x+4y

4y=300-3x

y=75-3/4x

Substitute the equation for y into the function for area.

A=x(75-3/4x)=75x-3/4x²

Find the derivative of the equation for area.

A'=75-1.5x

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 pig pen).

0=75-1.5x

x=75/1.5=50

A'is positive when x<50 and A' is negative when x>50 , therefore meaning that x=50 is a maximum. Since this value is a maximum, the area is maximized when the total length is 50 m.

Find the height ( y ) when x=50

y=75-3/4x=75-3/4(50)=37.5 m

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

50m by 37.5 m

A=50x37.5=1875 m²