The owner of the Rancho Grande has 3,060 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions (in yd) of the largest area he can enclose?

1170450 yd^2

Step-by-step explanation:

The first thing is to calculate the necessary perimeter, which would be like this:

2 * a + b = 3060

if we solve for b, we are left with:

b = 3060-2 * a

Now for the area it would be:

A = a * b = a * (3060-2 * a )

A = 3060 * a -2 * a ^ 2

To maximize the area, we calculate the derivative with respect to "a":

dA / da = d [3060 * a -2 * a ^ 2

]/gives

dA / day = 3060 - 4 * a

If we equal 0:

0 = 3060 - 4 * a

4 * a = 3060

a = 3060/4

a = 765 and d

Therefore b:

b = 3060 - 2 * a = 3060 - 1530 = 1530

A = a * b

A = 765 * 1530

A = 1170450 and d ^ 2

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.