Dividing by variable when solving algebraic equation

[robvalue]

I like questions like that though, because they show a simple application of maths to a real situation where guesswork would only get you so far.

For those that are interested, this is how you get the answer:

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You have 100m of fence to make a rectangle against a wall, which forms one of the four sides. So if the length of the rectangle coming away from the wall is x, you'll be using x also for the side above it, leaving 100 - 2x for the third side, opposite the wall.

So the area is

A = x(100 - 2x) = 100x - 2x^2

To find the maximum, we differentiate with respect to x, to find the rate of change in the area. To differentiate Ax^n, you get Anx^(n-1)

dA/dx = 100 - 4x

At a maximum or minimum amount, the rate of change is zero. (Think of the top or bottom of a quadratic curve; the tangent becomes flat.)

0 = 100 - 4x

x = 25

To check this is indeed a maximum, we take the second derivative:

d2A/dx^2 = -4

This means it is a maximum, as its less than zero. The rate of change is lessening as you approach and go over the point. Less than zero is a minimum.

So with x = 25, the side opposite the wall is 100 - 2*25 = 50

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Once you get to the formula for A in terms of x, there's an alternate method that doesn't involve using calculus. You can complete the square instead, and then pick the relevant value for x to give what is obviously the maximum value for A.