Farmer Ed has 300 meters of fencing, and wants to enclose a rectangular plot that borders on a river.
maximize the area. What is the largest area that can be enclosed?
Farmer Ed does not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

L = 150 meters

B = 75 meters

Area = 11250 square meters

Step-by-step explanation:

Step 1 :

Total length of fencing = 300 meters

Let L be the length and W be the width of the plot. The river side is not fenced.

Hence we have

L + 2W = 300 => L = 300-2W

Step 2 :

Area of the plot = length * width

=> A = L *W =  (300-2W) * W = 300B - 2W²

Step 3:

For  the area to be maximum, we need to have the first derivative to be 0 and the second derivative to be less than 0

The first derivative for the area equal to 0 is

300-4W = 0 => 4W = 300 = > W = 75

The second derivative for the area is -4 which is always less than 0.

Hence we have the maximum area when the width is 75 meters.

Step 4 :

When W = 75, the length

L = 300 - 2 * 75 = 300 -150 = 150

So when the length is 150 meters and the width is 75 meters , the area of the plot is maximized

The maximum area that can be enclosed is 75 * 150 = 11250 meters

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24