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

Guest Apr 19, 2022

#1**+1 **

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.

MaxWong Apr 19, 2022