Answer
Relative minimum at the points (1,1) and (-1,-1)
Work Step by Step
Step 1: First Partial Derivatives Given the function: \[f(x, y) = x^2 + y^2 + xy^2\] Find the first partial derivatives of \(f(x, y)\): \[f_x(x, y) = \frac{\partial}{\partial x}[x^2 + y^2 + xy^2] = 2x - x^2y^2\] \[f_y(x, y) = \frac{\partial}{\partial y}[x^2 + y^2 + xy^2] = 2y - xy^2\] Step 2: Second Partial Derivatives Find the second partial derivatives of \(f(x, y)\) and \(f_x(x, y)\): \[f_{xx}(x, y) = \frac{\partial^2}{\partial x^2}[2x - x^2y^2] = 2 - 2x^3y^2\] \[f_{yy}(x, y) = \frac{\partial^2}{\partial y^2}[2y - xy^2] = 2 - 2x^2y\] \[f_{xy}(x, y) = \frac{\partial^2}{\partial y \partial x}[2x - x^2y^2] = 2xy^2 - 2xy\] Critical Points To find the critical points, we set both first partial derivatives equal to zero: For \(f_x(x, y)\): \[2x - x^2y^2 = 0\] For \(f_y(x, y)\): \[2y - xy^2 = 0\] Solving these equations, we find two critical points: 1. \(x = 1, y = 1\) 2. \(x = -1, y = -1\) Second-Partial Test Now, we evaluate the second-partial test using the Hessian determinant \(D(x, y)\): \[D(x, y) = \left[f_{xx}(x, y) \cdot f_{yy}(x, y)\right] - \left[f_{xy}(x, y)\right]^2\] For the critical point (1,1): \[D(1, 1) = \left[(2 - 2(1)^3(1)^2) \cdot (2 - 2(1)^2(1))\right] - \left[(2(1)(1)^2 - 2(1)(1))\right]^2 = 32 > 0\] For the critical point (-1,-1): \[D(-1, -1) = \left[(2 - 2(-1)^3(-1)^2) \cdot (2 - 2(-1)^2(-1))\right] - \left[(2(-1)(-1)^2 - 2(-1)(-1))\right]^2 = 32 > 0\] Since \(D\) is positive for both critical points and \(f_{xx}\) is positive, we have relative minima at the points (1,1) and (-1,-1). So, the results are: Relative minimum at the points (1,1) and (-1,-1).
