Maximum and Minimum value

Given $x^{2} + y^{2} = 25$ , by the Cauchy-Schwarz inequality, $(x^{2} + y^{2})(1^{2} + 1^{2}) \ge (x+y)^{2}$ .

Then, $(25)(2) \ge (x+y)^{2}$ . Therefore, $(x+y) \le \sqrt{50} \approx \boxed{7.07}$ .

Given that $x^{2} + y^{2} = 25$ we can parameterize $x,y$ as $x = 5\cos(\theta), y = 5\sin(\theta), 0 \le \theta \le 2\pi$ . Then

$x + y = 5\cos(\theta) + 5\sin(\theta)) = 5\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\cos(\theta) + \dfrac{1}{\sqrt{2}}\sin(\theta)\right) = 5\sqrt{2}(\sin(45^{\circ})\cos(\theta) + \cos(45^{\circ})\sin(\theta)) = 5\sqrt{2}\sin(\theta + 45^{\circ})$ .

As $-1 \le \sin(\theta + 45^{\circ}) \le 1$ we see that the maximum possible value of $x + y$ is $5\sqrt{2} \approx \boxed{7.07}$ .

This problem can also be solved via Lagrange Multipliers.

Since x x + y y = 25, rewrite x+y as x + sqrt(25-x x). So let us state that f(x) = x + sqrt(25-x x). For either a maximum or minimum value of f(x) to occur df(x)/dx = 0. df(x)/dx = 1-x/sqrt(25-x^2) = 0 occurs when 25-x^2 = x^2 or when x = +/- 5/sqrt(2). y is hence equal to +/- 5/sqrt(2). The maximum value of x+y is hence 10/sqrt(2) = 7.07. We can also investigate the second derivative when x and y are equal to 5/sqrt(2) Second derivative is equal to -25/(25-x^2)^(3/2) and is < 0 which means that x + y has a local maximum at x and y equal to 5/sqrt(2).