```Question 120787
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Example 6.

A rancher has 600 meters of fence to enclose a rectangular corral with another fence dividing it in the middle as in the diagram below.

As indicated in the diagram, the four horizontal sections of fence will each be x meters long and the three vertical sections will each be y meters long.

The rancher's goal is to use all of the fence and enclose the largest possible area.

The two rectangles each have area xy, so we have

total area: A = 2xy.

There is not much we can do with the quantity A while it is expressed as a product of two variables. However, the fact that we have only 1200 meters of fence available leads to an equation that x and y must satisfy.

3y + 4x = 1200.

3y = 1200 - 4x.

y = 400 - 4x/3.

We now have y expressed as a function of x, and we can substitute this expression for y in the formula for total area A.

A = 2xy = 2x (400 -4x/3).

We need to find the value of x that makes A as large as possible. A is a quadratic function of x, and the graph opens downward, so the highest point on the graph of A is the vertex. Since A is factored, the easiest way to find the vertex is to find the x-intercepts and average.

2x (400 -4x/3) = 0.

2x = 0 or 400 -4x/3 = 0.

x = 0 or 400 = 4x/3.

x = 0 or 1200 = 4x.

x = 0 or 300 = x.

Therefore, the line of symmetry of the graph of A is x = 150, the average of 0 and 300.

Now that we know the value of x corresponding to the largest area, we can find the value of y by going back to the equation relating x and y.

y = 400 - 4x/3 = 400 -4(150)/3 = 200.

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