To make things more clear, think about the graph of P(x, y) = xy. You want to find the highest value on the graph of P(x, y) such that x + y = 80. With this constraint, we have a simple relationship between x and y, and they can be written in terms of each other. Namely, x = 80 - y or y = 80 - x.
Since we're working on this constraint and one-variable functions are easier to work with anyway, we can rewrite P(x, y) as P(x) = x(80 - x). This function has pretty straightforward graph, but with a significant feature: at some x value, P(x) actually stops getting bigger and becomes smaller instead. Meaning the slope changes from positive to negative. Intuitively (or by the Mean Value Theorem), there must be a point in this transition in which the slope is zero (right before it becomes negative). So we simply solve for what x makes the slope equal to zero: P'(x) = 0 => (-x2 + 80x)' = 0 => -2x + 80 = 0 => 80 = 2x => x = 40. So it looks like at x = 40, the slope is 0, and by our previous logic, this must be the point at which P(x)'s graph is the highest (i.e. the maximum). Now that we know x is equal to 40, it isn't hard to solve x + y = 80 for y.