let

x = shorter side of the fenced area that is the width of the rectangle

y = the length of the rectangle

the perimeter of the rectangle is 2x + 2y but since one side is along the highway (let that side by y) and is not fenced we have

2x + y = 400 => y = 400 - 2x

since the area of rectangle = width * length we can write for the area of the field

A(x) = x * (400 - 2x)

A(x) = - 2x^2 + 400x

we know that the quadratic y = ax^2 + bx + c with a<0 has maximum value at its vertex equal to c - (b^2/4a) so we have

A(x) = - 2x^2 + 400x has maximum value equal to 0 - ((400)^2/4*(-2)) = 20 000

by solving the quadratics

- 2x^2 + 400x = 20000

x = 100

y = 400 - 200 = 200

the dimensions of the field with maximum area are 100 and 200.

Lets consider the case when the highway is along x we will have

x + 2y = 400

y = 200 - (x/2)

A(x) = x*y

A(x) = x*(200 - (x/2))

A(x) = - x^2/2 + 200x

has maximum for x = 200 A(x) = 20,000 and y = 100

since x is the shorter side then x < y follows that this can not be the case.