Since we know that x + y = 1, then y = 1 - x

We can substitute this into P, so that it is now only a function of x (AKA "in terms of x"):

1. P = x² + x(1-x) - (1-x)²
2. P = x² + x - x² - (1 - 2x + x²)
3. P = x - 1 + 2x - x²
4. P = 3x - x² - 1

Then, to maximise a function, we find the derivative with respect to x (note that if we didn't substitute the y terms away, we would be in trouble here).

1. dP/dx = 3 - 2x
2. Stationary points occur when: dP/dx = 0
3. 3 - 2x = 0
4. x = 3/2 = 1.5

Also, since P is a parabola, we can do this question without differentiation calculus. We recall that the axis of symmetry in a parabola is equal to k = -b/2a, where y = ax² + bx + c

Since a = -1, and b = 3, then k = -3/(-2) = 3/2 = 1.5

The axis of symmetry is where the turning point resides, and as the parabola is negative (the coefficient to the x² term is -1), then we have an upside down parabola, which gives us a local maximum turning point!