AREA word problem that I can not solve :((

Question

A farmer is setting up separate pens for livestock by fencing off a rectangular area and running fence to create eight parallel pens. If the farmer has a total of 360 meters of fence, what is the largest area that can be contained?

I just don't understand the steps.

Answer 1

9x+2y=360

y = 180-9x/2

Area = A= xy =x(180-9x/2) = 180x-4.5x^2

A' = 180-9x = 0

x = 180/9 = 20

y =180- 4.5(20) = 90

max A = xy = 20(90) = 1800 square meters

Answer 2

Let's see now. If there are 8 parallel pens, the fencing involved, for a length x, is 9x for that direction of fence (there are 9 segments running that way), and (360 - 9x)/2 (the rest of the fencing, in 2 pieces!) for the perpendicular direction of fencing. The product (total area) = (360-9x)*x/2. Calculate that out, set first derivative with respect to x to zero (the function for area has a maximum, at which the slope of (area vs. x) graph is zero), and solve. So d(180x-(9/2)x^2)/dx = 180 - 9x . So 0 = 180 - 9x and x = 20, area = 20*90 = 1800 m^2 . Did that make it clearer?

-- Cheers, -- Mr. d.

Answer 3

Let's consider 8 parallel pens, dimension l and h. h=lx, so total area is 8 (lh) = 8xl^2

The perimeter of 8 parallel pens is 8h + 8h + 9l = 360; 16h +9l=360; l(16x+9) = 360

So l=360/(16x+9). Area A = 8xl^2 = 8 360^2 x /(16x+9)^2 max for x is when A' = 0

(16x+9) -x(16x+9) 2 16 so for x=9/16

l = 20 and A max = 8 x l^2 = 8 (20) 9/16 = 1800 m^2

Image at https://paragonmath.com/png/area_max.png