Getting familiar with the real problem

Question

In a village full of rivers and canals, to determine the length of the fence, we have the following parameters: we have 10 meters of fence and want to fence a rectangular field next to a straight river, using the river as one side of the fence. Since the river forms a natural boundary, we only need to fence the remaining three sides (two shorter sides and one longer side parallel to the river). What size will maximize the fenced area?

Answer 1

Let x = length of wire used for shorter sides (perpendicular to the river)

Let y = length of wire used for the longer side (assumed)--parallel to the river.

2x + y = 10 (constraint placed by given conditions).

So, y = 10 - 2x

The area of the rectangle is length times width:

A = xy

A = x(10 - 2x)

A(x) = -2x² + 10x

That quadratic function when graphed results in a parabola that opens downward. Locating the vertex gives the x value that generates the maximum area. There are several ways to locate the vertex. One way it to determine the axis of symmetry which passes through the vertex. For quadratic functions the axis of symmetry is given by the formula a = -b/2a where in this case b = 10 and a = -2.

So, x = -10/-4 = 5/2. This is the x-value that generates the maximum area.

y = 10 - 2(5/2) = 5

The dimensions that maximize the area are 2.5 by 5, and the maximum area is 25/2 or 12 1/2 square meters.

Since the roots of the quadratic function are 0 and 5 [from x (10 -2x)], the axis of symmetry can also be found by averaging those roots (0+5)/ 2 = 5/2.

Another way to determine the vertex is to complete the square to place the function in vertex form:

A = -2(x² - 5x + ??) -- where question marks complete the square

A = -2(x² - 5x + 25/4) + 25/2 -- The 25/4 completes the square on the trinomial, visualizing redistributing the -2 leads to the conclusion that -25/2 was added to the equation, so the +25/2 keeps it in balance.

A(x) = -2(x - 5/2)² + 25/2 -- vertex form (5/2, 25/2); rewrote perfect square trinomial as binomial squared