Let's solve the problem step by step! ## **1. Find the critical points of the functions** ### **a) \( f(x, y) = x^2 + 2y^2 - 4y + 6x \)** Find the critical points by setting the first partial derivatives to zero: \[ f_x = \frac{\partial f}{\partial x} = 2x + 6 \] \[ f_y = \frac{\partial f}{\partial y} = 4y - 4 \] Set \( f_x = 0 \) and \( f_y = 0 \): \[ 2x + 6 = 0 \implies x = -3 \] \[ 4y - 4 = 0 \implies y = 1 \] **Critical point:** \( (-3, 1) \) --- ### **b) \( g(x, y) = x^2 - 12xy + y \)** \[ g_x = \frac{\partial g}{\partial x} = 2x - 12y \] \[ g_y = \frac{\partial g}{\partial y} = -12x + 1 \] Set \( g_x = 0 \) and \( g_y = 0 \): \[ 2x - 12y = 0 \implies x = 6y \] \[ -12x + 1 = 0 \implies x = \frac{1}{12} \] Plug \( x = \frac{1}{12} \) into \( x = 6y \): \[ \frac{1}{12} = 6y \implies y = \frac{1}{72} \] **Critical point:** \( \left(\frac{1}{12}, \frac{1}{72}\right) \) --- ## **2. Use the Second Derivative Test** ### **a) For \( f(x, y) \):** \[ f_{xx} = 2, \quad f_{yy} = 4, \quad f_{xy} = 0 \] The discriminant: \[ D = f_{xx} f_{yy} - (f_{xy})^2 = (2)(4) - (0)^2 = 8 \] - \( D > 0 \) and \( f_{xx} > 0 \): **Local minimum.** ### **b) For \( g(x, y) \):** \[ g_{xx} = 2, \quad g_{yy} = 0, \quad g_{xy} = -12 \] The discriminant: \[ D = g_{xx} g_{yy} - (g_{xy})^2 = (2)(0) - (-12)^2 = -144 \] - \( D < 0 \): **Saddle point.** --- ## **3. Match with Graphs** - **A** shows a saddle surface. - **B** shows a parabolic bowl (minimum). So: - **\( f(x, y) \) matches (B)** (local minimum) - **\( g(x, y) \) matches (A)** (saddle point) --- ## **Summary Table** | Function | Critical Point | Nature | Graph | |------------------|----------------------------------|---------------|-------| | \( f(x, y) \) | \( (-3, 1) \) | Local Minimum | B | | \( g(x, y) \) | \( \left(\frac{1}{12}, \frac{1}{72}\right) \) | Saddle Point | A | --- ### **Final Answers** - \( f(x, y) \) has a local minimum at \( (-3, 1) \), matches graph **B**. - \( g(x, y) \) has a saddle point at \( \left(\frac{1}{12}, \frac{1}{72}\right) \), matches graph **A**.

Critical points are found where the first derivatives vanish, and their nature (minimum, maximum, saddle) is determined using the second derivative test via the discriminant \( D = f_{xx}f_{yy} - (f_{xy})^2 \). If \( D > 0 \) and \( f_{xx} > 0 \), the point is a local minimum; if \( D > 0 \) and \( \(f_{xx} < 0 \), it is a local maximum; if \( D < 0 \), the point is a saddle point.

Critical points occur where the first derivatives are zero, and their nature is determined by the second derivatives through the discriminant; positive discriminant indicates a minimum or maximum, while negative indicates a saddle point.