Finding Quadratic problems

Question

Farmer Ed has 2,000 meters of fencing, and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not fence the side along the river what is the largest area that can be enclosed?

Answer 1

For the equation of a parabola in the form y = ax² + bx + c, the max/min occurs at x = -b/2a.

 

Therefore from 

A = -2W² + 2000W, the area is maximum when W = -2000/-4, or 500.

Answer 2

Given dimension width (W) and length (L) we have the perimeter which has fence defined by:

P = 2*W + L = 2000 meters (assuming the other length is the river)

Then, we know that the area is given by:

A = L*W

 

To maximize the area, first define the function by one variable with the perimeter constraint (or limitation by fencing)

L = 2000 - 2*W

 

A = (2000 - 2*W)*W

A = 2*W*(1000 - W)

 

Note, this is a factored quadratic, so we can expand the terms:

A = -2W^2 + 2000W

 

To find the maximum of this function, since the quadratic faces downward (negative "a" term) we have:

vertex == Maximum point

 

The vertex is found by completing the square:

complete square = (1000/2)^2 = 250,000

added constant = -a*(1000/2)^2 = 500,000

A = -2*(W^2 - 1000W + 250000) + 500000

A = -2*(W-500)^2 + 500000

 

Vertex: (500,500000)

So the maximum area for the fencing is 500,000 square meters. The fencing dimensions are:

Width = 500; Length = 1000