Bunuel wrote:

The rectangle above is a diagram of the fencing surrounding an animal pen 20 meters by 5 meters. If an additional 4 square meters is to be enclosed by moving just one side of the pen, while retaining its rectangular shape, what is the least possible number of meters of additional fencing needed?

(A) 2.0

(B) 1.6

(C) 1.0

(D) 0.8

(E) 0.4

Attachment:

2017-09-20_1023_002.png

Area increases from 100 to 104 \(m^2\). One or both lengths must increase.

The "least possible number of meters of additional fencing needed" to obtain the additional area can be reframed: maximize the rectangle's area and minimize its perimeter.

The

least change in perimeter to obtain the maximum change in area will come from adding length to the shorter sides.

That is, we need add the fencing to the shorter sides with length 5, because:

---The minimum perimeter that yields maximum area for a rectangle is a square.

---Extend that principle: in order for an increase in area to be obtained with the least addition to the perimeter ("least number of meters of fencing"), this rectangle needs to get closer to a square in shape.

If we extend the longer side of 20, the rectangle becomes even more oblong = less like a square

Hence, take the shorter side with length 5. Both sides must be extended. Although both short sides are extended, to calculate area, only one of the two short sides is used.

Let \(x\) = additional meters of fencing needed

\(\frac{1}{2}x\) gets added to each side with length 5

5 + \(\frac{1}{2}x\) = New length of ONE short side, which gets multiplied by 20 to equal the new area

New area = \((20)(5 +\) \(\frac{1}{2}x\))

\(104 = 100 + 10x\)

\(4 = 10x\)

\(\frac{4}{10} = 0.4 = x\)

ANSWER E

P.S. By contrast, if you add \(\frac{1}{2} x\) to the (long) sides with length 20, the minimum possible additional meters of fencing = 1.6

_________________

At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123