We are given 600 m of fencing to enclose a rectangular area and are asked to maximize this area.

Let the dimensions of the rectangle be l (for length) and w (for width.)

Then the perimeter is l+w+l+w or 2l+2w. We know the perimeter is 600 m, so we have 2l+2w=600 or l+w=300. (Dividing both sides of the equation by 2.)

The area is given by A=l*w.

We now have two equations in two unknowns. l+w=300 and A=l*w, where we want to maximize the area.

Solving l+w=300 for w, we get w=300-l. We can then substitute this expression into the equation for the area. Thus A=l(300-l).

`A=300l-l^2` is a quadratic in l.

1. The graph of the quadratic is a parabola, opening down. (The leading coefficient, or the coefficient on the squared term, is negative.) So the graph has a maximum located at the vertex.

1a. If the quadratic is written in standard form, `f(x)=ax^2+bx+c`, then the vertex is located at `((-b)/(2a),f((-b)/(2a)))`. We can write `A(l)=-l^2+300l` so the first coordinate of the vertex is `(-300)/(2*(-1))=150`. Then the second coordinate is `A(150)=22500`. So the length is 150 m, the width is 300-150=150 m, and the area enclosed is 22,500 square meters.

1b. The vertex is located 1/2 way between the two real zeros. For A(l) the 2 zeros are (0,0) and (300,0). Thus the first coordinate is 1/2(300-0)=150, as above.

2. If you have calculus, we know the maximum occurs at a critical point of the function (the first derivative is zero or fails to exist).

If `A(l)=-l^2+300l` then `A'(l)=-2l+300`. Setting the derivative equal to zero we see -2l+300=0 ==> 2l=300 or l=150, as above.

The dimensions are 150 m by 150 m with an area of 22,500 square meters.