Question

Building sheep pens : darsh and darpan are fencing off a large rectangular area to build some temporary holding pens. to prep the males, females, and kids, they are separated into three smaller and equal-size pens partitioned

Answer 1

The larger rectangle will be a square with sides of 96 ft while the pens will have 96 ft of length and 31 ft of width

How to Maximize Area of rectangle?

The side lengths of the big rectangle are x and y and so we have the following expression for its area A:

A = x*y

This is tied to the constraint that the perimeter should not be larger than the available fencing , Thus;

2x + 2y = 384   (All the fencing is used to maximize the area)

y = (384 - 2x)/2

y = 192 - x

Put 192 - x for y in A to get;

A = x(192 - x)

A = 192x - x²

Completing the square gives;

A = 9216 - (x - 96)²

Thus;

A is maximum at;

x - 96 = 0

x = 96 ft

y = 192 - 96

y = 96 ft

Then since we maximised the area of the bigger rectangle , we have maximised the area of the smaller pens. The dimensions will be;

x_small = 96 ft /3

x_small = 31 ft

y_small = 96 ft

Complete question is;

Building sheep pens: It's time to drench the sheep again, so Chance and Chelsea-Lou are fencing off a large rectangular area to build some temporary holding pens. To prep the males, females, and lambs, they are separated into three smaller and equal-size pens partitioned within the large rectangle. If 384 ft of fencing is available and the maximum area is desired, what will be (a) the dimensions of the larger, outer rectangle? (b) the dimensions of the smaller holding pens?

Read more about maximizing area at; brainly.com/question/14853334