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mamaluj [8]
2 years ago
13

If 384 ft of fencing is available and the maximum area is desired, what will be the dimensions of the larger, outer rectangle? (

ii) what will be the dimensions of the smaller holding pens?
SAT
1 answer:
DIA [1.3K]2 years ago
6 0

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

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