Precalc question

Question

Jun has 300 meters of fencing to make a rectangular enclosure. She also wants to use some fencing to split the enclosure into two parts with a fence parallel to two of the sides. What dimensions should the enclosure have to have the maximum possible area?

Answer 1

The perimeter of a rectangular enclosure with sides w and l is equal to 2(W+L). Adding a length of the fence parallel to one of the sides to split the enclosure, the total is 2(W+L)+L=300.

The area of a rectangular enclosure is W*L. To find the dimensions that would maximize the area, express one variable through the other:

2W+3L=300

2W=300-3L

W=(300-3L)/2

Substitute in the area formula:

W*L=[(300-3L)*L]/2

[(300-3L)*L]/2=150L-1.5L²=-1.5L²+150L

It is a downward concave parabola and its vertex is its maximum, hence the maximum area can be achieved with the L that is the coordinate of the vertex.

x-coordinate of the vertex in a quadratic equation ax²+bx+c=0 is equal to (-b)/2a.

Therefore,

L=-150/(-3)=50

W=(300-3L)/2=(300-150)/2=75

The target dimensions are 75 by 50.

P.S. It does not matter whether the extra side splitting the enclosure into two parts is W or L.