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Chapter 13: Problem 24

Allow inequality constraints, optional but good. Find the minimum of \(f=3 x+5 y\) with the constraints \(g=\) \(x+2 y=4\) and \(h=x \geqslant 0\) and \(H=y \geqslant 0,\) using equations like (7). Which multiplier is zero?

Short Answer

Expert verified
The minimum is 10 at \((0, 2)\) with multiplier for \( x \geq 0 \) being zero.

Step by step solution

01

Understand the Problem

We need to find the minimum value of the function \( f = 3x + 5y \) subject to the constraints \( g = x + 2y = 4 \) and the inequality constraints \( x \geq 0 \) and \( y \geq 0 \). Lagrange multipliers will be used to find the minimum, particularly identifying when a multiplier is zero.
02

Set Up the Lagrangian

The Lagrangian function \( L \) is given by \( L = f + \lambda g \), where \( \lambda \) is the Lagrange multiplier. Thus, derive the Lagrangian as: \[ L = 3x + 5y + \lambda(4 - x - 2y). \]
03

Compute Partial Derivatives

To find critical points, compute the partial derivatives of the Lagrangian with respect to \( x \), \( y \), and \( \lambda \). Set them equal to zero:1. \( \frac{\partial L}{\partial x} = 3 - \lambda = 0 \).2. \( \frac{\partial L}{\partial y} = 5 - 2\lambda = 0 \).3. \( \frac{\partial L}{\partial \lambda} = x + 2y - 4 = 0 \).
04

Solve the System of Equations

From the first two equations, express \( \lambda \) from the first equation: \[ \lambda = 3. \]Substitute into the second equation:\[ 5 - 2\times 3 = 0 \Rightarrow \lambda = 2.5 eq \lambda \]. However, for equality, \( 2\lambda = 5 \), leads to a contradiction, therefore check boundary conditions of inequalities.
05

Check Boundary Conditions

Based on \( x = 0 \) or \( y = 0 \), we can find solutions. Solve the equations using the constraints:1. If \( x = 0 \), solve \( 0 + 2y = 4 \Rightarrow y = 2 \). Point \((0, 2)\).2. If \( y = 0 \), solve \( x = 4 \). Point \((4, 0)\).Evaluate both points on \( f(x, y) \):\( f(0, 2) = 5 \times 2 = 10 \), \( f(4, 0) = 3 \times 4 = 12 \).
06

Conclusion and Check Multipliers

The minimum value found is 10 at point \((0, 2)\). Here \( x = 0 \) indicates the multiplier (related to constraint \( x \geq 0 \)) results in zero, hence, this constraint is active.

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Key Concepts

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

Inequality Constraints
In optimization problems, constraints come in two flavors: equality and inequality. An inequality constraint limits the values that variables can take, ensuring they stay within a certain range, such as not going below zero.
A typical example is when we want values that satisfy conditions like:
  • \( x \geq 0 \)
  • \( y \geq 0 \)
Here's why inequality constraints are essential:
  • They define feasible regions for solutions.
  • They restrict variables to realistic conditions, such as preventing negative lengths or amounts.
In our problem, we apply inequality constraints to ensure variables \(x\) and \(y\) are non-negative, fulfilling real-world scenarios.
Optimization Problem
When dealing with optimization problems, the goal is to find the best solution from a set of possibilities. We often aim to minimize (or maximize) a particular function by tweaking the variables within permitted conditions.
The steps to solve an optimization problem usually involve:
  • Defining a target function to optimize, like \( f = 3x + 5y \).
  • Specifying the constraints, i.e., \( x + 2y = 4 \) for equality, and non-negativity constraints for inequalities.
  • Using techniques such as the method of Lagrange multipliers to work around the constraints.
The ultimate aim is to determine variable values where the target function achieves its minimum value, as illustrated with values \(x = 0\) and \(y = 2\). This strategy checks all conditions, finding the feasible and most efficient solution.
Linear Programming
Linear Programming (LP) is a powerful mathematical method used in optimization. It is perfect for problems where the objective and constraints can be expressed linearly.
The essentials of linear programming include:
  • Formulating a linear objective function to either maximize or minimize, like \( f = 3x + 5y \).
  • Using linear constraints, both equalities like \( x + 2y = 4 \), and inequalities such as \( x \geq 0 \), \( y \geq 0 \).
The main advantage of LP is its ability to handle complex problems systematically, often with large data sets. By finding intersections of constraints, solutions like point \((0, 2)\) are identified as optimal. LP ensures all solutions are realistic and bound by specified needs.

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Most popular questions from this chapter

Find the maximum of \(f=x y\) on the circle \(g=x^{2}+y^{2}=2\) by solving \(f_{x}=\lambda g_{x}\) and \(f_{y}=\lambda g_{y}\) and substituting \(x\) and \(y\) into f. Draw the level curve \(f=f_{\max }\) that touches the circle.

If \(f(x, y)=x y\) and \(x=u(t)\) and \(y=v(t),\) what is \(d f_{i} d t ?\) Probably all other rules for derivatives follow from the chain rule.

Find \(f_{x}, f_{y}, f_{x x}, f_{x y}, f_{y y}\) at the basepoint. Write the quadratic approximation to \(f(x, y)-\) the Taylor series through second order terms: \(f=e^{x+y}\) at (0,0)

Find the velocity \(v\) and the tangent vector \(T\). Then compute the rate of change \(d f / d t=\operatorname{grad} f \cdot \mathbf{v}\) and the slope \(d f / d s=\operatorname{grad} f \cdot \mathbf{T}\). \(f=x y \quad x=t^{2}+1 \quad y=3\)

A rectangle has sides on the \(x\) and \(y\) axes and a corner on the line \(x+3 y=12\). Find its maximum area.
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