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Let x and y be real numbers such that 2(x^2 + y^2) = x + y + 8. Find the maximum value of x - y.

Apr 19, 2022

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Moving some terms yields: $$2x^2 + 2y^2 - x - y - 8 = 0$$

Note that geometrically, this is the equation of a circle, because we can rewrite the equation as follows:

$$\begin{array}{rcl} 2x^2 + 2y^2 - x - y - 8 &=& 0\\ x^2 + y^2 - \dfrac12 x - \dfrac12 y - 4&=&0\\ \left(x - \dfrac14\right)^2 + \left(y - \dfrac14\right)^2 &=& \dfrac{33}8 \end{array}$$

Consider the straight line x - y = c, where c is varying through real numbers.

You can find the maximum value of c such that the line and the circle intersects, i.e., $$\begin{cases}2x^2 + 2y^2 - x - y - 8 = 0\\x - y = c​​​​\end{cases}$$ has solutions.

To do so, you need to substitute y = x - c into the circle equation, then consider the discriminant of the resulting equation. Since the system has solutions, the discriminant of the resulting equation must be nonnegative. Solving the equation $$\Delta \geq 0$$ will give you the suitable range of values of c.

Please tell me if you want more explanations or you need more help.

Apr 19, 2022