The gradient of the level curve f(x,y) can be found using partial derivatives.
[tex]\Delta f = (\frac{\delta f}{\delta x},\frac{\delta f}{\delta y}) = (1,1)[/tex]
This means that the max value lies on the vector <1,1> which is equivalent to the line y = x.
Find the points along edge of disc where y = x.
[tex]x^2 + y^2 = 1, y = x \\ \\ x^2 = \frac{1}{2} \\ \\ x = \pm \frac{\sqrt{2}}{2} [/tex]
These 2 solutions provide both max and min values.
Let x be positive will give max value for f(x,y)
[tex]f(x,y)_{max} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + 4 = \sqrt{2} + 4 [/tex]
