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Chapter 9: Problem 5

Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$ f(x, y)=x+y, \quad x^{2}+y^{2}=1 $$

Short Answer

Expert verified
Max: \(\sqrt{2}\), Min: \(-\sqrt{2}\) at respective points on the unit circle.

Step by step solution

01

Identify the Objective and Constraint Functions

The objective function is given by \( f(x, y) = x + y \). The constraint is \( x^2 + y^2 = 1 \). Our goal is to maximize or minimize the objective function subject to this constraint.
02

Formulate the Lagrangian

Form the Lagrangian function by incorporating the constraint: \[ L(x, y, \lambda) = x + y + \lambda (1 - x^2 - y^2) \]where \( \lambda \) is the Lagrange multiplier.
03

Take Partial Derivatives

Find the partial derivatives of the Lagrangian with respect to \( x, y, \) and \( \lambda \): - \( \frac{\partial L}{\partial x} = 1 - 2\lambda x = 0 \)- \( \frac{\partial L}{\partial y} = 1 - 2\lambda y = 0 \)- \( \frac{\partial L}{\partial \lambda} = 1 - x^2 - y^2 = 0 \) (which is the original constraint).
04

Solve the System of Equations

From step 3, solve the equations:1. \( 1 = 2\lambda x \) implying \( x = \frac{1}{2\lambda} \),2. \( 1 = 2\lambda y \) implying \( y = \frac{1}{2\lambda} \),3. \( x^2 + y^2 = 1 \).Substitute \( x = \frac{1}{2\lambda} \) and \( y = \frac{1}{2\lambda} \) into the constraint:\( \left( \frac{1}{2\lambda} \right)^2 + \left( \frac{1}{2\lambda} \right)^2 = 1 \).Simplifying yields:\( \frac{1}{4\lambda^2} + \frac{1}{4\lambda^2} = 1 \)\( \frac{1}{2\lambda^2} = 1 \)\( 2\lambda^2 = 1 \)\( \lambda^2 = \frac{1}{2} \).
05

Determine Values of \(x\) and \(y\)

Solving \( \lambda^2 = \frac{1}{2} \), we have \( \lambda = \pm \frac{1}{\sqrt{2}} \). When \( \lambda = \frac{1}{\sqrt{2}} \), \( x = \sqrt{2}/2 \) and \( y = \sqrt{2}/2 \), and similarly, when \( \lambda = -\frac{1}{\sqrt{2}} \), \( x = -\sqrt{2}/2 \) and \( y = -\sqrt{2}/2 \).
06

Evaluate \(f(x, y)\) at these points

Calculate the objective function values:- At \( (x, y) = (\sqrt{2}/2, \sqrt{2}/2) \), \( f(x, y) = \sqrt{2}/2 + \sqrt{2}/2 = \sqrt{2} \).- At \( (x, y) = (-\sqrt{2}/2, -\sqrt{2}/2) \), \( f(x, y) = -\sqrt{2}/2 - \sqrt{2}/2 = -\sqrt{2} \).
07

Conclude Maximum and Minimum Values

The maximum value of \( f(x, y) \) is \( \sqrt{2} \) at \( (\sqrt{2}/2, \sqrt{2}/2) \), and the minimum value is \( -\sqrt{2} \) at \( (-\sqrt{2}/2, -\sqrt{2}/2) \).

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Key Concepts

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

Objective Function
In optimization problems, the objective function is crucial as it represents the quantity that needs to be maximized or minimized. In this exercise, our objective function is given by \( f(x, y) = x + y \). This function is the key to defining what we are trying to optimize.
When working with objective functions, it is important to:
  • Clearly define the function with respect to the variables involved.
  • Understand what the function represents in the context of the problem. Here, it represents a simple linear combination of \( x \) and \( y \).
The ultimate goal of the problem is to find values of \( x \) and \( y \) that maximize or minimize this objective function while also respecting certain constraints.
Constraint Function
Constraints are conditions that must be met for solutions to be valid. In this case, the constraint is given by the equation \( x^2 + y^2 = 1 \). This describes a circle with a radius of 1, and it is essential because it limits where we can look for solutions.
Working with constraint functions involves understanding:
  • The nature of the constraint, in this case, a circle defined by a quadratic equation.
  • How it restricts the set of possible solutions, meaning any solution must lie on this circle.
  • The role of constraints in ensuring that solutions are realistic and adhere to the problem's requirements.
The presence of this constraint makes the problem one of constrained optimization, for which Lagrange multipliers provide an elegant solution method.
Partial Derivatives
Partial derivatives are a fundamental concept when working with functions of multiple variables. In this exercise, we use them to understand how changes in \( x \) and \( y \) affect the Lagrangian function. The partial derivatives are calculated as follows:
  • \( \frac{\partial L}{\partial x} = 1 - 2\lambda x \)
  • \( \frac{\partial L}{\partial y} = 1 - 2\lambda y \)
  • \( \frac{\partial L}{\partial \lambda} = 1 - x^2 - y^2 \)
These derivatives are critical for setting up the system of equations to solve. By setting them to zero, we find the stationary points of the Lagrangian, which potentially provide the maximum or minimum values of the objective function under the given constraint.
System of Equations
Once we have the partial derivatives, the next step is to solve the resulting system of equations. These equations stem from setting the partial derivatives to zero, which is essential for finding optimal points. In this case, the system consists of:
  • \( 1 = 2\lambda x \)
  • \( 1 = 2\lambda y \)
  • \( x^2 + y^2 = 1 \)
By solving this system, we determine possible values for \( x \), \( y \), and \( \lambda \), which satisfy both the original function and the constraint condition. Solving these equations leads us to identify the maximum and minimum values of our objective function under the given constraint, guiding us to the optimal solution of the problem.

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Most popular questions from this chapter

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)-x^{2}-2 x y+3 y^{2}-8 y $$

A company has the production function \(P(x, y)\), which gives the number of units that can be produced for given values of \(x\) and \(y\); the cost function \(C(x, y)\) gives the cost of production for given values of \(x\) and \(y\). (a) If the company wishes to maximize production at a cost of $$\$ 50,000,$$ what is the objective function \(f\) ? What is the constraint equation? What is the meaning of \(\lambda\) in this situation? (b) If instead the company wishes to minimize the costs at a fixed production level of 2000 units, what is the objective function \(f ?\) What is the constraint equation? What is the meaning of \(\lambda\) in this situation?

Are about the money supply, \(M\), which is the total value of all the cash and checking account balances in an economy. It is determined by the value of all the cash, \(B\), the ratio, \(c\), of cash to checking deposits, and the fraction, \(r\), of checking account deposits that banks hold as cash: $$M=\frac{c+1}{c+r} B$$ (a) Find the partial derivative. (b) Give its sign. (c) Explain the significance of the sign in practical terms. $$ \partial M / \partial B $$

The sales of a product, \(S=f(p, a)\), are a function of the price, \(p\), of the product (in dollars per unit) and the amount, \(a\), spent on advertising (in thousands of dollars). (a) Do you expect \(f_{p}\) to be positive or negative? Why? (b) Explain the meaning of the statement \(f_{a}(8,12)=\) 150 in terms of sales.

The quantity, \(Q\), of a certain product manufactured depends on the quantity of labor, \(L\), and of capital, \(K\), used according to the function $$Q=900 L^{1 / 2} K^{2 / 3}$$ Labor costs $$\$ 100$$ per unit and capital costs $$\$ 200$$ per unit. What combination of labor and capital should be used to produce 36,000 units of the goods at minimum cost? What is that minimum cost?
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