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Use Lagrange multipliers to maximize each function \(f(x, y)\) subject to the constraint. (The maximum values do exist.) \(f(x, y)=3 x y, \quad x+3 y=12\)

Short Answer

Expert verified
The maximum value is 36 at \((x, y) = (6, 2)\).

Step by step solution

01

Define the Problem and the Constraint

We want to maximize the function \(f(x, y) = 3xy\) subject to the constraint \(x + 3y = 12\). The method of Lagrange multipliers involves introducing an auxiliary variable (the Lagrange multiplier) \(\lambda\) and setting up the Lagrangian function.
02

Formulate the Lagrangian

Construct the Lagrangian \( \mathcal{L}(x, y, \lambda) = 3xy + \lambda(12 - x - 3y) \). Here, \(\lambda\) is the Lagrange multiplier.
03

Find Partial Derivatives

Compute the partial derivatives of \( \mathcal{L} \) with respect to \(x, y,\) and \(\lambda\):1. \( \frac{\partial \mathcal{L}}{\partial x} = 3y - \lambda = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 3x - 3\lambda = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 12 - x - 3y = 0 \) Each of these derivatives is set to zero to find the critical points.
04

Solve the Equations

From \( \frac{\partial \mathcal{L}}{\partial x} = 3y = \lambda \) and \( \frac{\partial \mathcal{L}}{\partial y} = 3x = 3\lambda \), solve for \(x\) and \(y\):- \( \lambda = 3y \)- \( \lambda = x \)Equating the two results: \(3y = x\).Substitute \(x = 3y\) into the constraint \(x + 3y = 12\):- \(3y + 3y = 12\)- \(6y = 12\)- \(y = 2\)Using \(y = 2\), substitute back to find \(x\):- \(x = 3(2) = 6\).
05

Verify the Maximum Value

With \((x, y) = (6, 2)\), check that this satisfies the constraint \(x + 3y = 12\). Plug these values back into the function:- \(f(6, 2) = 3 \cdot 6 \cdot 2 = 36\).Thus, the maximum value of \(f(x, y)\) is 36 when \((x, y) = (6, 2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Multivariable Calculus
Multivariable calculus extends calculus principles to functions of multiple variables. It is crucial for analyzing scenarios where more than one input influences an outcome. A central component is differentiating and integrating these functions to observe changes in relation to one another.
Here, the function we aim to maximize is of two variables, namely \(x\) and \(y\). This means our calculus strategies move from one-dimensional (like \(f(x)\)) to higher dimensions.
Key elements in multivariable calculus include:
  • Partial Derivatives: These allow us to observe changes in the function with respect to one variable, keeping others constant.
  • Gradient: A vector consisting of partial derivatives that point toward the direction of greatest increase of the function.
By turning our attention to these aspects, we gather insights into the behavior and tendencies of multivariable functions such as \(f(x, y) = 3xy\).
Exploring Constraint Optimization
Constraint optimization is about finding the best outcome of a function while adhering to restrictions on the variables. In this context, we have a constraint: \(x + 3y = 12\). This constraint limits the possible values \(x\) and \(y\) can assume to ensure the constraint is satisfied.
The approach employs a technique called the method of Lagrange multipliers. By introducing a new variable, \(\lambda\), into our equations, we construct the Lagrangian, \(\mathcal{L}(x, y, \lambda) = 3xy + \lambda(12 - x - 3y)\).
This method is powerful because:
  • It transforms a constrained problem into an unconstrained one by adding a term to the original function that incorporates the constraint.
  • Setting the partial derivatives of the Lagrangian to zero helps locate points that maximize or minimize the function subject to the constraint.
By balancing the primary function and constraint, the method of Lagrange multipliers allows us to determine optimal solutions not easily reachable otherwise.
Unpacking Partial Derivatives
Partial derivatives are crucial in multivariable calculus. These derivatives describe how a multivariable function changes as one variable is altered, keeping others constant. For our Lagrangian, \(\mathcal{L}(x, y, \lambda) = 3xy + \lambda(12 - x - 3y)\), partial derivatives are computed with respect to \(x\), \(y\), and \(\lambda\).
Calculating each derivative:
  • \(\frac{\partial \mathcal{L}}{\partial x} = 3y - \lambda = 0\) points towards finding how the Lagrangian changes concerning \(x\).
  • \(\frac{\partial \mathcal{L}}{\partial y} = 3x - 3\lambda = 0\) provides insight into changes in \(y\).
  • \(\frac{\partial \mathcal{L}}{\partial \lambda} = 12 - x - 3y = 0\) captures how changes in constraint satisfaction affect the solution.
Solving these equations gives us the critical points where the function could achieve maximum or minimum values. This critical assessment, enabled through partial derivatives, helps ensure that our solution adheres to all conditions inherently posed by the functions and equations involved.

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Most popular questions from this chapter

THE SLOPE OF \(f(x, y)=c\) On page 556 we used the fact that the slope in the \(x\) - \(y\) plane of the curve defined by \(f(x, y)=c\) (for constant \(c\) ) is given by the formula \(-\frac{f_{x}}{f_{v}}\). Verify this formula by justifying the following five steps. a. If \(f(x, y)=c\) can be solved explicitly for a function \(y=F(x)\), then we may write \(f(x, F(x))=c .\) Justify: \(f(x+\Delta x, F(x+\Delta x))-f(x, F(x))=0\). b. Justify: \(f(x+\Delta x, F(x+\Delta x)-F(x)+F(x))\) \(-f(x, F(x))=0\) c. Defining \(\Delta F\) by \(\Delta F=F(x+\Delta x)-F(x)\), we may write the previous equation as $$ f(x+\Delta x, \Delta F+F(x))-f(x, F(x))=0 $$ Then, writing \(F\) for \(F(x)\), this becomes $$ f(x+\Delta x, F+\Delta F)-f(x, F)=0 $$ Justify: \(f_{x} \Delta x+f_{y} \Delta F \approx 0\). d. Justify: \(\frac{\Delta F}{\Delta x} \approx-\frac{f_{x}}{f_{y}}\). e. Justify: \(\frac{d F}{d x}=-\frac{f_{x}}{f_{y}}\). This shows that the slope of \(F(x)\), and therefore the slope of \(f(x, y)=c\), is \(-\frac{f_{x}}{f_{y}}\).

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