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Given that x + y = 1, the maximum value of P = x2 + xy - y2 occurs for x when?

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2 Answers

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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"):


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


dP/dx = 3 - 2x
Stationary points occur when: dP/dx = 0
3 - 2x = 0
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!

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For more general problems of this type try Lagrange Multipliers

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