I found: ##A=250xx500=125000m^2##

Considering the field as: I know that the perimiter (only on 3 sides) to be fenced is equal to the meters of fence at disposal of the farmer: ##2h+b=1000m## (1) The area will be ##A=bxxh## (2)

From (1) ##b=1000-2h## in (2)

##A=(1000-2h)xxh=1000h-2h^2##

Derive ##A## with ##h##:

##A’=1000-4h## equal it to zero to maximize it:

##1000-4h=0##

##h=1000/4=color(red)(250m)## use this back in (1) you find ##b=color(red)(500m)##: Use these dimensions in (2): ##A=250xx500=color(blue)(125000m^2)##

I found: ##A=250xx500=125000m^2##

Considering the field as: I know that the perimiter (only on 3 sides) to be fenced is equal to the meters of fence at disposal of the farmer: ##2h+b=1000m## (1) The area will be ##A=bxxh## (2)

From (1) ##b=1000-2h## in (2)

##A=(1000-2h)xxh=1000h-2h^2##

Derive ##A## with ##h##:

##A’=1000-4h## equal it to zero to maximize it:

##1000-4h=0##

##h=1000/4=color(red)(250m)## use this back in (1) you find ##b=color(red)(500m)##: Use these dimensions in (2): ##A=250xx500=color(blue)(125000m^2)##