A farmer with 800 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fen

A farmer with 800 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens.|||let x and y = dimension of each pen


800 = 8y + 5x


5x = 800 - 8y


x = 160 - 8y/5





A = 4xy


A = 4(160-8y/5)y


A = 640y - 32y虏/5


dA/dy = 640 - 64y/5


640 = 64y/5


y = 50 ft


x = 160 - 8(50)/5


x = 80 ft





total Area = 4(50)(80)


total Area = 16000 ft虏


Area of each pen = 50(80) = 4000 ft虏|||Let the main rectangle be dimensions X and Y, with X%26gt;=Y without loss of generality.





To divide into four pens they will need 3 more piece of fencing Y long. [You could do it X but that would require more fencing to cover the same area]





So the problem is:





Maximize X*Y


Subject to X+X+Y+Y+3Y=800 or 2X+5Y=800





Rewrite 2X+5Y=800 as





X=400-(5/2)Y





Substitute





XY=[(400-(5/2)Y]Y=400Y-(5/2)Y^2





Maximize 400Y-(5/2)Y^2





So take the first derivative of the equation and set it equal to zero





d[400Y-(5/2)Y^2]/dy = 0


400-2(5/2)Y=0


400-5Y=0


Y=80





X=(400)-(5/2)Y=400-(5/2)(80)=400-200=2...





So the area is 200*80=16000sqft





Fence used [as a check] is





200+200+80+80+80+80+80=800