A rectangular field runs along a river requires fencing on only 3 sides. What is the maximum area possible wit?

A rectangular field runs along a river requires fencing on only 3 sides. What is the maximum area possible with 500 meters of fencing?





Please explain this in algebraic form and in detail, because I am lost. Thx!





The answer is supposed to be: 31250m^2.|||500 meters of fencing on 3 sides:





x: width of rectangle in meters


y: height of rectangle in meters





lets say the fence runs twice the height and once the length (otherwise just define height/width the other way around ;)





so 500 = x + 2y





you're looking for the biggest area possible. the area is





A(x,y) = x*y.





to find a maximum/minimum of a function, take it's derivative and set it to equal zero.





A'(x,y) = 0





since this is a function in x and y, it's much easier if you first get rid of one of the variables, and then derive, do that with the help of your boundary condition from above





500 = x + 2y


x = 500 - 2y





therefore





A(y) = (500 - 2y)*y = 500y - 2y^2





derive to y and set to equal zero





A'(y) = 500 - 4y = 0


4y = 500


y = 125 m





x = 500 - 2y = 250 m.





The area then is





A = 125m*250m = 31250 m^2|||Area = A = L*W


P = 2*w + L


P = 500 m


500 - 2w = L





Area = (500 - 2w)*w


A = 500w - 2w^2


A = -2(w^2 - 250w )


A = -2(w^2 - 250w + (250/2)^2 ) + 2*125^2





A = - 2(w- 125)^2 + 2*125^2





The width is 125


The length is 250





P = 2w + L = 2*125 + 250 = 500 which is the check.|||500= 2w + L


area = w * L


area = w(500-2w)


area = 500w - 2w^2


d/dw(area) = 500 - 4w = 0 =%26gt; 500 = 4w =%26gt; w = 125


500-250 = L = 250





area = 250(125) = 31250m^2


w = 125


L = 250