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?

a) the larger rectangle will be actually a square with sides of 96 ft

b) the pens will have 96 ft of length and 31 ft of width

Step-by-step explanation:

denoting the side lengths of the big rectangle as x and y , we have the following expression for its area A:

A= x*y

that is tied to the constraint that the perimeter should not be larger than the available fencing , thus

2*x + 2*y = 384   (we use all the fencing to maximise the area)

y = (384 -2*x)/2 = 192 - x

replacing in A

A= x*y = x*(192 - x) = 192*x - x²

we can complete the square to rearrange the equation of A

A=192*x - x² + 96² - 96² = 96² - (  x² - 192*x + 96²) = 9216 - (x - 96)²

then A is maximum for x - 96 = 0 → x=96 ft

thus y= 192 - 96 = 96 ft

thus the rectangle with maximum area is actually a square with sides 96 ft ( if a rectangle is required, then diminish any of the lengths by the smallest amount possible. For example x= 95.99999 and y=96.00001)

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 = 31 ft , y small = 96 ft

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right