One side of a rectangular field is adjacent to a river. there are 160 yards of fencing available for the other three sides. what dimensions of the field will result in the maximum area?

The dimensions of 40 yards by 80 yards of the field will result in maximum area of 3200 yards squared.

Step-by-step explanation:

Let "x" and "y" be the lengths of the sides of rectangle.

So, the perimeters of three sides is:

2x + y = 160 yards

y = 160 - 2x     eqn (1)

Area = A = xy = x(160 - 2x)

A(x) = 160x - 2x²     [in terms of x]

Taking 1st and setting value to zero:

A'(x) = 0 = 160 - 4x

4x = 160

x = 40 yards

Now, taking 2nd derivative to check whether value is minimum or maximum.

A"(x) = -4

since, A"(x) < 0

therefore, it means the value x = 40 yards is maximum.

Thus, from eqn (1):

y = 160 - 2(40 yards)

y = 80 yards

and maximum area will be:

Amax = (40 yards)(80 yards)

Amax = 3200 yards squared

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.