Question

A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 548 feet of fencing is used. How do you find the dimensions of the playground that maximize the total enclosed area?

Answer 1

Answer: 274/3 and 137 ft

Step-by-step explanation:

Total fence used: 548 ft

So, perimeter will be 548 ft, including the fence that divides the land.

Consider the playground the drawing below.

__________x________

|                                        |

|                                        |   y

|                                        |

|__________________|

Let's consider that the playground is going to be divided parallel to y (and there's no problem, because if you choose parallel to x, the area will be the same). So, the perimeter (the sum of all sides) will be:

2x + 3y = 548

And the area

A = x.y

Isolating x, we have:

x = (548 - 3y)/2

Substituting the x in the Area equation:

A = (548 - 3y)/2 . y = 548y - 3y²/2 = 274y - 3/2y²

A = 274y - 3/2y²

So, it's a quadratic function. And, to find out the maximum area that the playground can have, we need to find the y-vertex of the parabola: y-vertex = Δ/4a

a = -3/2    b = 274     c = 0

Δ = 274² - 4.(-3/2).0 = 274² = 75076

Amax = -Δ/4a = -75076/4.(-3/2) = 75076/6 = 12512.7 ft²

y = -b/2a = -274/2.(-3/2) = 274/3 ft

x = (548 - 3y)/2 = (548 - 3.274/3)/2 = 274/2 = 137 ft