Afarmer has 336 feet of fencing to enclose 2 adjacent rectangular pig pens sharing a common side. what dimensions should be used for each pig pen so that the enclosed area will be a maximum? the two adjacent pens have the same dimensions.

x  =  84 ft  the largest side of the pens

y  =  56 ft the common side is that length

Step-by-step explanation:

Let call  x the total length of one side, and y the other side, and  the common side such as:

p(perimeter) = 336  = 2x + 3y           y  =  ( 336 - 2x ) / 3

And the area of the whole area

A(t)  =  x * y        A(x)  = x* (336  - 2x)/3      A(x)  = [336x  - 2x² ]/3

Taking derivatives both sides of the equation

A´(x)   =[ ( 336 -  4x )*3]/9         ⇒   A´(x)   =  0      ( 336 -  4x )*3  = 0

336 - 4x = 0

x  =  336/4

x  =  84 ft         and      y  = ( 336 - 2x ) / 3        y = ( 336 - 168)/3

y  =  56 ft

hope this .

r3t40

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!