The dimensions of the larger outer rectangle and smaller holding pens are respectively; 96ft x 96ft and 31ft x 96ft
Let the side lengths of the big rectangle be x and y. Thus the area of the big triangle is; A = xy
If the fencing available is 384 ft, then perimeter is 384 ft. Thus;
2x + 2y = 384
y = (384 - 2x)/2
y = 192 - x
Putting 192 - x for y in A gives us;
A = x(192 - x) = 192x - x²
Completing the square to rearrange the equation of A gives us;
A = 9216 - (x - 96)²
If A is to be a maximum, then;
x - 96 = 0
x = 96 ft
y = 192 - 96 = 96 ft
Since we maximised the area of the bigger rectangle , we have maximised the area of the smaller pens. The dimensions of the smaller holding pens will be;
x' = 96 ft /3 = 31 ft
y' = 96 ft
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