Let's break down the problem and solve it step-by-step: --- ### **Step 1: Understanding the Problem** - The dashed curve represents the constraint \( g(x, y) = k \). - The solid curves are level curves of the function \( f(x, y) \). - The arrows indicate the gradient \( \nabla f \) at several points. - **Question:** Where does the absolute maximum of \( f \) under the constraint \( g(x, y) = k \) occur: at P, Q, R, or somewhere else? --- ### **Step 2: The Method of Lagrange Multipliers** The absolute maximum (or minimum) of \( f(x, y) \) under the constraint \( g(x, y) = k \) occurs where the gradients of \( f \) and \( g \) are parallel, i.e., \[ \nabla f = \lambda \nabla g \] This means the gradients of \( f \) and \( g \) point in the same or opposite directions at the maximum/minimum. --- ### **Step 3: Analyzing the Diagram** - At **P**, the gradient \( \nabla f \) is **not** perpendicular to the dashed curve (the constraint). - At **Q**, the gradient \( \nabla f \) is **perpendicular** to the dashed curve. - At **R**, the gradient \( \nabla f \) is **not** perpendicular to the dashed curve. **Note:** The gradient of a function is always perpendicular to its level curve. So, for the maximum or minimum under the constraint, we want the gradient of \( f \) to be perpendicular to the constraint curve (which means they are parallel to the gradient of \( g \)). --- ### **Step 4: Conclusion** - **Q** is the point where \( \nabla f \) is perpendicular to the constraint curve \( g(x, y) = k \). - Therefore, the absolute maximum (or minimum, depending on the direction of the gradients and level curves) of \( f \) under the constraint occurs at **Q**. --- ### **Final Answer** **The absolute maximum of \( f \) under the constraint \( g(x, y) = k \) occurs at Q.** --- **Explanation:** At Q, the gradient of \( f \) is perpendicular to the constraint curve. This is the necessary condition for a maximum or minimum under a constraint, as established by the method of Lagrange multipliers. Points P and R do not satisfy this condition, so they cannot be extrema under the constraint.

Let's approach the problem systematically, incorporating detailed reasoning and calculations where applicable. --- ### **Understanding the problem** - The **dashed curve** represents the **constraint** \( g(x,y) = k \). - The **solid curves** are **level curves** of the function \( f(x,y) \). - The **arrows** indicate the **gradient vectors** \( \nabla f \) at various points. - The goal is to find **where** the **absolute maximum** of \( f(x,y) \) occurs **under the constraint** \( g(x,y) = k \). The options are points \( P \), \( Q \), \( R \), or elsewhere. --- ### **Step 1: Recall the method of Lagrange multipliers** The **necessary condition** for an extremum of \( f \) subject to the constraint \( g(x, y) = k \) is: \[ \nabla f(x, y) = \lambda \nabla g(x, y) \] This implies: - The **gradients are parallel**. - The **direction** of \( \nabla f \) aligns with \( \nabla g \). **Geometrically:** - **Level curves** of \( f \) are orthogonal to \( \nabla f \). - **Constraint curve** \( g(x,y) = k \) is orthogonal to \( \nabla g \). At an extremum under the constraint, the **gradient of \( f \)** is **parallel** to the **gradient of \( g \)**, meaning: \[ \nabla f \perp \text{tangent to the constraint curve} \] or equivalently, \[ \nabla f \parallel \nabla g \] --- ### **Step 2: Analyze the diagram** From the diagram: - **At Q**, \( \nabla f \) appears **perpendicular** to the dashed curve (the constraint). - **At P** and **R**, the gradient \( \nabla f \) is **not** perpendicular to the constraint. **Important note:** - Since \( \nabla f \) is **perpendicular** to the **level curve** of \( f \), - and the **constraint curve** \( g(x,y)=k \) has gradient \( \nabla g \), - the **condition for extrema** is: \[ \nabla f \parallel \nabla g \] which geometrically means: \[ \text{level curve of } f \quad \text{is tangent to} \quad g(x,y) = k \] and at such points, \( \nabla f \) and \( \nabla g \) are **parallel**. --- ### **Step 3: Interpretation of the diagram** - At **Q**, the gradient \( \nabla f \) is **perpendicular** to the constraint curve, indicating: \[ \nabla f \perp \text{curve} \] - The **constraint curve** is also orthogonal to \( \nabla g \). - Since \( \nabla f \) is perpendicular to the constraint at **Q**, and the **gradients** are **parallel** at the extremum, **Q** satisfies the necessary condition: \[ \nabla f \parallel \nabla g \] **Therefore, the point \( Q \) is where the maximum (or minimum) occurs.** --- ### **Step 4: Confirming the maximum** - The **direction** of \( \nabla f \) at \( Q \) points **away** from the level curves, and since \( \nabla f \) is perpendicular to the level curve at \( Q \), this aligns with the idea of an extremum at that point. - The **gradient vectors** (shown as arrows) at \( Q \) are **aligned** with the gradient of the constraint \( g \), confirming the necessary condition. --- ### **Step 5: Final conclusion** - The point where the **absolute maximum** of \( f \) under the constraint occurs is **at \( Q \)**. --- ### **Summary:** | Point | Gradient \( \nabla f \) | Relation to constraint \( g(x,y)=k \) | |---------|------------------------|-----------------------------------------| | **Q** | Perpendicular to level curve of \( f \), aligned with \( \nabla g \) | **Parallel** to \( \nabla g \) — satisfies the Lagrange multiplier condition — **candidate for extremum** | | **P** | Not perpendicular | Not satisfying the necessary condition | | **R** | Not perpendicular | Not satisfying the necessary condition | --- ## **Final Answer:** \[ \boxed{ \textbf{The absolute maximum of } f \text{ under the constraint occurs at point } \quad Q. } \] --- ### **Additional note:** - The **Lagrange multiplier method** confirms that extrema occur where \( \nabla f \) and \( \nabla g \) are **parallel**. - The **diagram** visually indicates this at **Q**, where the gradient vectors align appropriately. **Hope this detailed explanation clarifies the reasoning!**