Let's analyze the function step-by-step: ### **Given Function:** \[ f(x, y) = 27x^3 - 8y^3 + 36xy^2 - 54x^2y - 27x^2 - 12y^2 + 36xy + 9x - 6y + 11 \] Defined on \(\{(x, y) : x \leq \frac{1}{3}, y \geq 0 \}\). --- ## **Step 1: Find the Stationary (Critical) Points** Stationary points are where the gradient is zero: \[ \frac{\partial f}{\partial x} = 0, \quad \frac{\partial f}{\partial y} = 0 \] ### **Compute Partial Derivatives** #### **With respect to \(x\):** \[ \frac{\partial f}{\partial x} = 81x^2 + 36y^2 - 108xy - 54x + 36y + 9 \] #### **With respect to \(y\):** \[ \frac{\partial f}{\partial y} = -24y^2 + 72xy - 54x^2 - 24y + 36x - 6 \] --- ## **Step 2: Set the Derivatives Equal to Zero** ### \[ \boxed{ \begin{align*} 81x^2 + 36y^2 - 108xy - 54x + 36y + 9 &= 0 \tag{1} \\ -24y^2 + 72xy - 54x^2 - 24y + 36x - 6 &= 0 \tag{2} \end{align*} } \] Let's try to find solutions. #### **Try \(x = \frac{1}{3}\), \(y = 0\) (since these are the boundaries):** Plug into (1): \[ 81 \left(\frac{1}{3}\right)^2 + 36(0)^2 - 108 \left(\frac{1}{3}\right)(0) - 54 \left(\frac{1}{3}\right) + 36(0) + 9 = 81 \cdot \frac{1}{9} - 18 + 9 = 9 - 18 + 9 = 0 \] Plug into (2): \[ -24(0)^2 + 72 \left(\frac{1}{3}\right)(0) - 54 \left(\frac{1}{3}\right)^2 - 24(0) + 36 \left(\frac{1}{3}\right) - 6 = -54 \cdot \frac{1}{9} + 12 - 6 = -6 + 12 - 6 = 0 \] So, \(\left(x, y\right) = \left(\frac{1}{3}, 0\right)\) is a stationary point. --- #### **Try \(x = -1\), \(y = 1\):** Plug into (1): \[ 81(-1)^2 + 36(1)^2 - 108(-1)(1) - 54(-1) + 36(1) + 9 = 81 + 36 + 108 + 54 + 36 + 9 = 324 \] Not zero. #### **Try \(x = 1\), \(y = 1\):** Plug into (1): \[ 81(1)^2 + 36(1)^2 - 108(1)(1) - 54(1) + 36(1) + 9 = 81 + 36 - 108 - 54 + 36 + 9 = 0 \] Plug into (2): \[ -24(1)^2 + 72(1)(1) - 54(1)^2 - 24(1) + 36(1) - 6 = -24 + 72 - 54 - 24 + 36 - 6 = 0 \] So, \((1, 1)\) is another stationary point. But, recall the domain: \(x \le \frac{1}{3}\), \(y \ge 0\). So \((1, 1)\) is **not in domain**. Let's try \(x = 0\): Plug into (1): \[ 0 + 36y^2 - 0 - 0 + 36y + 9 = 36y^2 + 36y + 9 \] Set equal to 0: \[ 36y^2 + 36y + 9 = 0 \implies 4y^2 + 4y + 1 = 0 \implies (2y + 1)^2 = 0 \implies y = -\frac{1}{2} \] But \(y \ge 0\), not possible. Try \(y = 0\): Plug into (2): \[ -0 + 0 - 54x^2 - 0 + 36x - 6 = -54x^2 + 36x - 6 = 0 \] \[ 54x^2 - 36x + 6 = 0 \] \[ 9x^2 - 6x + 1 = 0 \] \[ x = \frac{6 \pm \sqrt{36 - 36}}{18} = \frac{6}{18} = \frac{1}{3} \] So, \(x = \frac{1}{3}, y = 0\) is stationary point (already found). --- #### **Conclusion: Only one stationary point** \((x, y) = (\frac{1}{3}, 0)\) in the domain. --- ## **Step 3: Evaluate the Statements** ### (1) \(f\) has two stationary points. **False** (Only one in the domain). ### (2) 11 is a local maximum of \(f\). At \((0, 0)\), \(f(0,0) = 11\), but \((0,0)\) is **not a critical point**, as shown above. ### (3) \(f\) has a local maximum at each critical point. Only one critical point, check if it is a maximum. ### (4) There is a critical point at which \(f\) does not have a local maximum. If the only critical point is not a maximum, then (4) is correct. ### (5) At each stationary point, the Hessian matrix of \(f\) is negative definite. Let's check the Hessian at \((\frac{1}{3}, 0)\): --- ## **Step 4: Calculate the Hessian at \((\frac{1}{3}, 0)\)** Second derivatives: \[ f_{xx} = 162x - 108y - 54 \] \[ f_{yy} = -48y + 72x - 24 \] \[ f_{xy} = f_{yx} = 72y - 108x + 36 \] At \((x, y) = (\frac{1}{3}, 0)\): \[ f_{xx} = 162 \cdot \frac{1}{3} - 0 - 54 = 54 - 54 = 0 \] \[ f_{yy} = 0 + 72 \cdot \frac{1}{3} - 24 = 24 - 24 = 0 \] \[ f_{xy} = 0 - 108 \cdot \frac{1}{3} + 36 = -36 + 36 = 0 \] So the Hessian matrix is the zero matrix: \[ H = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \] This is **not negative definite**. --- ## **Step 5: Final Answer** **The only critical point is at \((\frac{1}{3}, 0)\), and the Hessian is the zero matrix (not negative definite), and so we cannot classify it as a maximum. Thus, statement (4) is correct:** \[ \boxed{(4)\ \text{There is a critical point at which } f \text{ does not have a local maximum.}} \]

Let's analyze the problem carefully step-by-step, including all calculations and reasoning: --- ### **Given Function:** \[ f(x, y) = 27x^3 - 8y^3 + 36xy^2 - 54x^2 y - 27x^2 - 12 y^2 + 36 xy + 9x - 6 y + 11 \] **Domain:** \(\{(x,y) : x \leq \frac{1}{3}, y \geq 0\}\) --- ## **Step 1: Find the critical points (stationary points)** Stationary points occur where the gradient vanishes: \[ \frac{\partial f}{\partial x} = 0, \quad \frac{\partial f}{\partial y} = 0 \] ### **Compute partial derivatives** **Partial derivative with respect to \(x\):** \[ f_x = \frac{\partial}{\partial x} \left(27 x^3 - 8 y^3 + 36 xy^2 - 54 x^2 y - 27 x^2 - 12 y^2 + 36 xy + 9x - 6 y + 11 \right) \] Term-by-term: - \(27 x^3 \Rightarrow 81 x^2\) - \(-8 y^3 \Rightarrow 0\) - \(36 xy^2 \Rightarrow 36 y^2\) - \(-54 x^2 y \Rightarrow -108 x y\) - \(-27 x^2 \Rightarrow -54 x\) - \(-12 y^2 \Rightarrow 0\) - \(36 xy \Rightarrow 36 y\) - \(9x \Rightarrow 9\) - \(-6 y + 11 \Rightarrow 0\) **Therefore:** \[ \boxed{ f_x = 81 x^2 + 36 y^2 - 108 x y - 54 x + 36 y + 9 } \] --- **Partial derivative with respect to \(y\):** Term-by-term: - \(27 x^3 \Rightarrow 0\) - \(-8 y^3 \Rightarrow -24 y^2\) - \(36 xy^2 \Rightarrow 72 x y\) - \(-54 x^2 y \Rightarrow -54 x^2\) - \(-27 x^2 \Rightarrow 0\) - \(-12 y^2 \Rightarrow -24 y\) - \(36 xy \Rightarrow 36 x\) - \(9x \Rightarrow 0\) - \(-6 y \Rightarrow -6\) - \(11 \Rightarrow 0\) **Thus:** \[ \boxed{ f_y = -24 y^2 + 72 x y - 54 x^2 - 24 y + 36 x - 6 } \] --- ## **Step 2: Find critical points by solving \(f_x=0\) and \(f_y=0\).** --- ### **Equation (1):** \[ 81 x^2 + 36 y^2 - 108 x y - 54 x + 36 y + 9 = 0 \] ### **Equation (2):** \[ -24 y^2 + 72 x y - 54 x^2 - 24 y + 36 x - 6 = 0 \] --- ## **Step 3: Search for solutions** ### **Observation:** Test some boundary points, especially where the derivatives simplify. --- ### **Try \((x, y) = (\frac{1}{3}, 0)\):** Calculate \(f_x\): \[ f_x = 81 (\frac{1}{3})^2 + 36 \cdot 0^2 - 108 \cdot \frac{1}{3} \cdot 0 - 54 \cdot \frac{1}{3} + 36 \cdot 0 + 9 \] \[ = 81 \cdot \frac{1}{9} + 0 - 0 - 18 + 0 + 9 = 9 - 18 + 9 = 0 \] Calculate \(f_y\): \[ f_y = -24 \cdot 0^2 + 72 \cdot \frac{1}{3} \cdot 0 - 54 \left(\frac{1}{3}\right)^2 - 24 \cdot 0 + 36 \cdot \frac{1}{3} - 6 \] \[ = 0 + 0 - 54 \cdot \frac{1}{9} + 0 + 12 - 6 = -6 + 12 - 6 = 0 \] **Conclusion:** \((x, y) = (\frac{1}{3}, 0)\) **is a critical point**. --- ### **Check if this point is within the domain:** \[ x = \frac{1}{3} \leq \frac{1}{3} \quad \text{(on the boundary)}, \quad y=0 \geq 0 \] **Yes, it is within the domain.** --- ### **Are there other critical points?** Try \(x=0\): Calculate \(f_x\): \[ 81 \cdot 0^2 + 36 y^2 - 108 \cdot 0 \cdot y - 54 \cdot 0 + 36 y + 9 = 36 y^2 + 36 y + 9 \] Set to zero: \[ 36 y^2 + 36 y + 9 = 0 \implies 4 y^2 + 4 y + 1=0 \] \[ (2 y + 1)^2 = 0 \implies y = -\frac{1}{2} \] But \(y \ge 0\), so discard. Similarly, check \(y=0\): Calculate \(f_y\): \[ -24 \cdot 0 + 0 - 54 x^2 - 0 + 36 x - 6 = -54 x^2 + 36 x - 6 \] Set to zero: \[ -54 x^2 + 36 x - 6=0 \implies 54 x^2 -36 x +6=0 \] Divide by 6: \[ 9 x^2 - 6 x + 1=0 \] Solutions: \[ x= \frac{6 \pm \sqrt{36 - 36}}{18} = \frac{6}{18} = \frac{1}{3} \] which is the same critical point \((x,y) = (\frac{1}{3}, 0)\). --- ## **Summary of critical points:** - **Only one critical point in the domain: \(\boxed{\left(\frac{1}{3}, 0\right)}\).** --- ## **Step 4: Classify the critical point** ### **Calculate second derivatives to form Hessian** \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 162 x - 108 y -54 \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = -48 y + 72 x - 24 \] \[ f_{xy} = f_{yx} = 72 y - 108 x +36 \] --- ### **Evaluate at \(\left(\frac{1}{3}, 0\right)\):** \[ f_{xx} = 162 \cdot \frac{1}{3} - 108 \cdot 0 - 54 = 54 - 0 - 54= 0 \] \[ f_{yy} = -48 \cdot 0 + 72 \cdot \frac{1}{3} - 24 = 0 + 24 - 24= 0 \] \[ f_{xy} = 72 \cdot 0 - 108 \cdot \frac{1}{3} + 36 = 0 -36 +36= 0 \] **Hessian matrix:** \[ H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \] - The Hessian is **zero** (neither positive nor negative definite). --- ## **Step 5: Analyze the nature of the critical point** - Since the Hessian is zero, the second derivative test is inconclusive. - To understand whether it is a maximum, minimum, or saddle, evaluate \(f\) at the critical point and nearby points. --- ### **Value of \(f\) at \(\left(\frac{1}{3}, 0\right)\):** \[ f\left(\frac{1}{3}, 0\right) = 27 \left(\frac{1}{3}\right)^3 - 8 \cdots + \text{(rest of terms)} \] Calculate each term: - \(27 \times \frac{1}{27} = 1\) - Remaining terms: \[ -8 \times 0 + 36 \times \frac{1}{3} \times 0 - 54 \times \left(\frac{1}{3}\right)^2 \times 0 - 27 \times \left(\frac{1}{3}\right)^2 - 12 \times 0 + 36 \times \frac{1}{3} \times 0 + 9 \times \frac{1}{3} - 6 \times 0 + 11 \] Simplify: \[ = 1 + 0 + 0 - 27 \times \frac{1}{9} + 0 + 0 + 3 + 0 + 11 \] \[ = 1 - 3 + 3 + 11 = (1 - 3) + (3 + 11) = -2 + 14 = 12 \] **Value of \(f\) at the critical point: \(\boxed{12}\).** --- ### **Check \(f\) nearby:** - At \((x,y) = ( \frac{1}{3} + 0.01, 0)\): \[ x = 0.3433, \quad y=0 \] Calculate \(f\): \[ f(0.3433, 0) \approx 27 \times 0.3433^3 + \text{other terms} \] The leading term: \[ 27 \times 0.3433^3 \approx 27 \times 0.0407 \approx 1.1 \] which is close to the value at the critical point (12). Similar checks indicate the critical point is not a maximum or minimum but possibly a saddle or flat point. --- ## **Final conclusions:** - **Only one critical point** exists in the domain: \(\left(\frac{1}{3}, 0\right)\). - **Hessian matrix** at this point is zero (not negative definite). - The critical point is **not necessarily a maximum**; the second derivative test is inconclusive. --- ## **Matching with the options:** 1. **f has two stationary points.** **→ False** (only one). 2. **11 is a local maximum of f.** Evaluate at \((0,0)\): \[ f(0,0) = 11 \] But \((0,0)\) is **not a critical point**. So, **False**. 3. **f has a local maximum at each critical point.** Only one critical point, and the Hessian is zero, so **not necessarily** a maximum. **False**. 4. **There is a critical point at which f does not have a local maximum.** Given the Hessian is zero, and the value at the critical point is not a maximum. **True**. 5. **At each stationary point, the Hessian is negative definite.** Hessian is zero, not negative definite. **False**. --- ## **Final answer:** **\(\boxed{\textbf{(4)}\ \text{There is a critical point at which } f \text{ does not have a local maximum.}}\)**