Find the critical points of the function f(x,y) = 1 - cuberoot(x2 + y2).

Step 1: Find the first-order partial derivatives f_x and f_y. The given function is f(x,y) = 1 - [3]x^2 + y^2. We can rewrite this as f(x,y) = 1 - (x^2 + y^2)^1/3. Differentiate f(x,y) with respect to x (treating y as a constant): f_x = ()/( x) (1 - (x^2 + y^2)^1/3) f_x = 0 - (1)/(3)(x^2 + y^2)^(1/3)-1 · ()/( x)(x^2 + y^2) f_x = -(1)/(3)(x^2 + y^2)^-2/3 · (2x) f_x = -(2x)/(3(x^2 + y^2)^2/3) Differentiate f(x,y) with respect to y (treating x as a constant): f_y = ()/( y) (1 - (x^2 + y^2)^1/3) f_y = 0 - (1)/(3)(x^2 + y^2)^(1/3)-1 · ()/( y)(x^2 + y^2) f_y = -(1)/(3)(x^2 + y^2)^-2/3 · (2y) f_y = -(2y)/(3(x^2 + y^2)^2/3) Step 2: Find the critical points. Critical points occur where f_x = 0 and f_y = 0, or where f_x or f_y are undefined. If we set f_x = 0: -(2x)/(3(x^2 + y^2)^2/3) = 0 This implies 2x = 0, so x = 0. If we set f_y = 0: -(2y)/(3(x^2 + y^2)^2/3) = 0 This implies 2y = 0, so y = 0. However, if x=0 and y=0, the denominators 3(x^2 + y^2)^2/3 become 3(0^2 + 0^2)^2/3 = 0. Division by zero means that f_x and f_y are undefined at (0,0). Therefore, (0,0) is the only critical point. Step 3: Classify the critical point (0,0). The second derivative test (D-test) cannot be used here because the second partial derivatives are also undefined at (0,0) due to the (x^2+y^2)^-5/3 term in their denominators. Instead, we analyze the behavior of the function directly around (0,0). The function is f(x,y) = 1 - [3]x^2 + y^2. Let r = sqrt(x^2+y^2). This represents the distance from the origin (0,0) to the point (x,y). Since r 0, we have r^2 = x^2+y^2 0. The term [3]x^2+y^2 = (x^2+y^2)^1/3 is always non-negative. The minimum value of (x^2+y^2)^1/3 occurs when x^2+y^2=0, which is at (0,0). At (0,0), (0^2+0^2)^1/3 = 0. So, f(0,0) = 1 - 0 = 1. For any other point (x,y) (0,0), we have x^2+y^2 > 0. This means (x^2+y^2)^1/3 > 0. Therefore, for any (x,y) (0,0), f(x,y) = 1 - (x^2+y^2)^1/3 < 1. Since f(x,y) < f(0,0) for all (x,y) (0,0), the critical point (0,0) is a local maximum. The value of the function at this local maximum is f(0,0) = 1. Summary of results: • Local minima: There are no local minima. • Local maxima: There is a local maximum at (0,0) with a value of f(0,0) = 1. • Saddle points: There are no saddle points. That's 2 down. 3 left today — send the next one.