Let's solve the problem **step by step** using the **method of Lagrange multipliers**. --- ### **Step 1: Set up the function and constraint** - Function to minimize: \( f(x, y, z) = x + y + z \) - Constraint: \( g(x, y, z) = (x + 1)^2 + y^2 + z^2 = 1 \) --- ### **Step 2: Lagrange Multipliers** Set up the Lagrangian: \[ \mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda \left( (x+1)^2 + y^2 + z^2 - 1 \right) \] --- ### **Step 3: Take Partial Derivatives** Set each partial derivative to 0: #### For \(x\): \[ \frac{\partial \mathcal{L}}{\partial x} = 1 - \lambda \cdot 2(x+1) = 0 \implies 1 = 2\lambda(x+1) \] #### For \(y\): \[ \frac{\partial \mathcal{L}}{\partial y} = 1 - \lambda \cdot 2y = 0 \implies 1 = 2\lambda y \] #### For \(z\): \[ \frac{\partial \mathcal{L}}{\partial z} = 1 - \lambda \cdot 2z = 0 \implies 1 = 2\lambda z \] #### For \(\lambda\): \[ \frac{\partial \mathcal{L}}{\partial \lambda} = - \left( (x+1)^2 + y^2 + z^2 - 1 \right) = 0 \] --- ### **Step 4: Solve the System** From above: - \( 1 = 2\lambda(x+1) \implies x+1 = \frac{1}{2\lambda} \implies x = \frac{1}{2\lambda} - 1 \) - \( 1 = 2\lambda y \implies y = \frac{1}{2\lambda} \) - \( 1 = 2\lambda z \implies z = \frac{1}{2\lambda} \) Plug into the constraint: \[ (x+1)^2 + y^2 + z^2 = 1 \] \[ \left( \frac{1}{2\lambda} \right)^2 + \left( \frac{1}{2\lambda} \right)^2 + \left( \frac{1}{2\lambda} \right)^2 = 1 \] \[ 3 \left( \frac{1}{2\lambda} \right)^2 = 1 \] \[ 3 \cdot \frac{1}{4\lambda^2} = 1 \] \[ \frac{3}{4\lambda^2} = 1 \] \[ 4\lambda^2 = 3 \implies \lambda^2 = \frac{3}{4} \implies \lambda = \pm\frac{\sqrt{3}}{2} \] --- ### **Step 5: Find the corresponding values** #### **Case 1: \(\lambda = \frac{\sqrt{3}}{2}\)** \[ x = \frac{1}{2\lambda} - 1 = \frac{1}{\sqrt{3}} - 1 \] \[ y = z = \frac{1}{2\lambda} = \frac{1}{\sqrt{3}} \] \[ f(x, y, z) = x + y + z = \left(\frac{1}{\sqrt{3}} - 1\right) + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = 3\left(\frac{1}{\sqrt{3}}\right) - 1 = \sqrt{3} - 1 \] #### **Case 2: \(\lambda = -\frac{\sqrt{3}}{2}\)** \[ x = \frac{1}{2\lambda} - 1 = -\frac{1}{\sqrt{3}} - 1 \] \[ y = z = -\frac{1}{\sqrt{3}} \] \[ f(x, y, z) = x + y + z = \left(-\frac{1}{\sqrt{3}}-1\right) -\frac{1}{\sqrt{3}} -\frac{1}{\sqrt{3}} = -3\left(\frac{1}{\sqrt{3}}\right) - 1 = -\sqrt{3} - 1 \] --- ### **Step 6: Minimum value** Among the values \(\sqrt{3} - 1\) and \(-\sqrt{3} - 1\), the **minimum** is: \[ \boxed{-\sqrt{3} - 1} \] However, check the options. The closest is **option (d): \(-2\sqrt{3} - 1\)**, but our calculation gives \(-\sqrt{3} - 1\). Let's recheck the calculations for possible missing negative sign or algebraic slip. --- #### **Recheck:** Actually, from the constraint: \((x+1)^2 + y^2 + z^2 = 1\) Let \(y = z = a\), \(x = b-1\), then: \[ (b)^2 + a^2 + a^2 = 1 \implies b^2 + 2a^2 = 1 \] But from the equations: \(x+1 = y = z = t\) So \(x + 1 = y = z = t\) So: \[ x = t - 1, \quad y = t, \quad z = t \] Constraint: \[ (t)^2 + (t)^2 + (t)^2 = 1 \implies 3t^2 = 1 \implies t^2 = \frac{1}{3} \implies t = \pm \frac{1}{\sqrt{3}} \] When \(t = \frac{1}{\sqrt{3}}\): \[ x = \frac{1}{\sqrt{3}} - 1, \quad y = \frac{1}{\sqrt{3}}, \quad z = \frac{1}{\sqrt{3}} \] Sum: \[ x + y + z = \left(\frac{1}{\sqrt{3}} - 1\right) + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = 3\left(\frac{1}{\sqrt{3}}\right) - 1 = \sqrt{3} - 1 \] When \(t = -\frac{1}{\sqrt{3}}\): \[ x = -\frac{1}{\sqrt{3}} - 1, \quad y = -\frac{1}{\sqrt{3}}, \quad z = -\frac{1}{\sqrt{3}} \] Sum: \[ x + y + z = \left(-\frac{1}{\sqrt{3}} - 1\right) -\frac{1}{\sqrt{3}} -\frac{1}{\sqrt{3}} = -3\left(\frac{1}{\sqrt{3}}\right) - 1 = -\sqrt{3} - 1 \] So the **minimum** is: \[ \boxed{-\sqrt{3} - 1} \] But this is not listed in the options, but the closest is **(d) \(-2\sqrt{3} - 1\)**. **However, based on the correct calculation, the answer should be:** \[ \boxed{-\sqrt{3} - 1} \] Since option (b) is \(-\sqrt{3} - 1\), **option (b) is correct**.

Let's carefully solve the problem step by step. --- ### **Problem Restatement:** Find the **minimum** of the function: \[ f(x, y, z) = x + y + z \] subject to the constraint: \[ (x + 1)^2 + y^2 + z^2 = 1 \] --- ### **Step 1: Recognize the problem** This is a constrained optimization problem. We are minimizing a linear function \(f\) over a sphere (since the constraint is a sphere equation). The method of Lagrange multipliers is suitable here. --- ### **Step 2: Set up the Lagrangian** Define the Lagrangian: \[ \mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda \left( (x+1)^2 + y^2 + z^2 - 1 \right) \] --- ### **Step 3: Find the partial derivatives** Set derivatives with respect to \(x, y, z\) to zero: \[ \frac{\partial \mathcal{L}}{\partial x} = 1 - \lambda \cdot 2(x+1) = 0 \] \[ \frac{\partial \mathcal{L}}{\partial y} = 1 - 2 \lambda y = 0 \] \[ \frac{\partial \mathcal{L}}{\partial z} = 1 - 2 \lambda z = 0 \] And the derivative with respect to \(\lambda\): \[ (x+1)^2 + y^2 + z^2 = 1 \] --- ### **Step 4: Solve the system** From the derivatives: \[ 1 = 2 \lambda (x+1) \Rightarrow x + 1 = \frac{1}{2 \lambda} \] \[ 1 = 2 \lambda y \Rightarrow y = \frac{1}{2 \lambda} \] \[ 1 = 2 \lambda z \Rightarrow z = \frac{1}{2 \lambda} \] Note that \( y = z \). Define \( t = y = z = \frac{1}{2 \lambda} \). Express \( x \) in terms of \(\lambda\): \[ x = \frac{1}{2 \lambda} - 1 = t - 1 \] --- ### **Step 5: Use the constraint to find \(\lambda\)** Substitute \( x = t - 1 \), \( y = t \), \( z = t \) into the constraint: \[ (x+1)^2 + y^2 + z^2 = 1 \] \[ (t)^2 + t^2 + t^2 = 1 \] \[ 3 t^2 = 1 \] \[ t^2 = \frac{1}{3} \] \[ t = \pm \frac{1}{\sqrt{3}} \] --- ### **Step 6: Find the corresponding \(x\)** For each \( t \): - When \( t = \frac{1}{\sqrt{3}} \): \[ x = t - 1 = \frac{1}{\sqrt{3}} - 1 \] - When \( t = - \frac{1}{\sqrt{3}} \): \[ x = - \frac{1}{\sqrt{3}} - 1 \] --- ### **Step 7: Compute \(f(x, y, z)\) for these points** Recall: \[ f = x + y + z = x + 2 t \] **Case 1: \( t = \frac{1}{\sqrt{3}} \)** \[ f = \left( \frac{1}{\sqrt{3}} - 1 \right) + 2 \times \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} - 1 + \frac{2}{\sqrt{3}} = \left( \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} \right) - 1 = \frac{3}{\sqrt{3}} - 1 \] \[ f = \sqrt{3} - 1 \] **Case 2: \( t = - \frac{1}{\sqrt{3}} \)** \[ f = \left( - \frac{1}{\sqrt{3}} - 1 \right) + 2 \times \left( - \frac{1}{\sqrt{3}} \right) = - \frac{1}{\sqrt{3}} - 1 - \frac{2}{\sqrt{3}} = \left( - \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}} \right) - 1 = - \frac{3}{\sqrt{3}} - 1 \] \[ f = - \sqrt{3} - 1 \] --- ### **### **Step 8: Determine the minimum** Between the two values: - \(\sqrt{3} - 1 \approx 1.732 - 1 = 0.732\) - \(- \sqrt{3} - 1 \approx -1.732 - 1 = -2.732\) The **minimum** value is: \[ \boxed{- \sqrt{3} - 1} \] --- ### **### **Final Answer:** \[ \boxed{- \sqrt{3} - 1} \] which corresponds to **option (b)**. --- **Summary:** - Using Lagrange multipliers, we found critical points. - Evaluated the function at these points. - The minimum value of \(f\) under the given constraint is **\(- \sqrt{3} - 1\)**, matching option (b).