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The dimensions of the smaller holding pens from the parameters given are; 96 ft and 31 ft
<h3>What dimensions will maximize the area?</h3>
From the complete question, if the side lengths of the big rectangle are x and y, then the expression for the area A is:
A = x*y
Then perimeter since we have 384 ft of fencing available is;
2x + 2y = 384
y = (384 - 2x)/2
y = 192 - x
Put 192 - x for y in area formula;
A = x(192 - x)
A = 192x - x²
Completing the square of this are equation gives;
A = 9216 - (x - 96)²
This means that A is maximum at x - 96 = 0
Thus, A is maximum when x = 96 ft
At A_max; y = 192 - 96 = 96 ft
Since the area of the bigger rectangle has been maximized, it means that we have also maximized the area of the smaller pens. Therefore its' dimensions will be;
x_small = 96 ft/3 = 31 ft
y_small = 96 ft
Read more about maximizing area at; brainly.com/question/9819619
Answer:
False is the correct answer.
Explanation:
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