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Chapter 5: Problem 1

Use Lagrange multipliers to optimize $$ 2 x^{2}-x y $$ subject to $$ x+y=12 $$

Short Answer

Expert verified
The optimized value of the function is -12 at \(x = 2\) and \(y = 10\).

Step by step solution

01

Set Up the Lagrangian

Define the Lagrangian function as \[\begin{equation} \ \ \ L(x,y,\lambda) = 2x^2 - xy + \lambda (x + y - 12) \ \ \ \end{equation}\]where \( \ \ \lambda \) is the Lagrange multiplier.
02

Compute Partial Derivatives

Compute the partial derivatives of the Lagrangian with respect to \( \ \ x, \ y, \) and \( \ \ \lambda \).\begin{align*} \ \ L_x &= \frac{\partial L}{\partial x} = 4x - y + \lambda = 0 \ \ \ \ \ L_y &= \frac{\partial L}{\partial y} = -x + \lambda = 0 \ \ \ \ \ L_\lambda &= \frac{\partial L}{\partial \lambda} = x + y - 12 = 0 \ \ \ \end{align*}
03

Solve the System of Equations

Solve the system of equations formed by setting the partial derivatives equal to zero.1) \( \ 4x - y + \lambda = 0 \) 2) \( \ -x + \lambda = 0 \) 3) \( \ x + y - 12 = 0 \) From equation 2, \( \ \lambda = x \). Substitute \( \ \lambda \) into equation 1 to get: \( \ 4x - y + x = 0 \) or \( \ 5x - y = 0 \), which implies \( \ y = 5x \). Substitute \( \ y = 5x \) into equation 3: \( \ x + 5x - 12 = 0 \) or \( \ 6x = 12 \), so \( \ x = 2 \). Finally, using \( \ y = 5x \ \), \( \ y = 10\ \).
04

Calculate the Objective Function

Substitute \( \ x = 2 \) and \( \ y = 10 \) into the objective function to find its optimized value: \[\begin{equation} \ \ \ 2x^2 - xy = 2(2)^2 - (2)(10) = 8 - 20 = -12 \ \ \end{equation}\]

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Key Concepts

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

Optimization
Optimization is all about finding the best solution to a problem. It involves maximizing or minimizing a particular function. For example, in business, you might want to maximize profit or minimize costs.
In mathematics, optimization revolves around finding the maximum or minimum values of a function. This function is also known as an objective function.
To optimize a function, you often need to use calculus and other mathematical tools. Lagrange multipliers are one such tool, particularly useful when dealing with constraints.
Constrained Optimization
Constrained optimization deals with optimizing a function while considering some restrictions or conditions (constraints).
In our exercise, we want to maximize or minimize the objective function \(2x^2 - xy\) while being restricted by the condition \(x + y = 12\).
This type of problem commonly appears in economics, engineering, and even everyday life scenarios.
The method of Lagrange multipliers is a powerful technique to solve these kinds of problems. It helps you find the optimal points where the function respects the constraints.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus. They show how a function changes as one of its input variables changes, while keeping the other variables constant.
For example, in our exercise, we compute the partial derivatives of the Lagrangian function with respect to \(x, y,\text{ and }\boldsymbol{位}\).
This step is crucial because setting these partial derivatives to zero gives us a system of equations. Solving this system helps find the optimal solution.
Partial derivatives are often symbolized using the 鈭� notation. For instance, \(\frac{\boldsymbol{鈭倉L}{\boldsymbol{鈭倉x}\text{ and }\frac{\boldsymbol{鈭倉L}{\boldsymbol{鈭倉y}\).
System of Equations
A system of equations is a set of multiple equations that you solve together. These equations share common variables, and solving the system means finding the values of these variables that satisfy all the equations simultaneously.
In our exercise, we form a system of three equations derived from partial derivatives and the constraint:
  • 4\boldsymbol{x} - \boldsymbol{y} + \boldsymbol{位} = 0
  • -\boldsymbol{x} + \boldsymbol{位} = 0
  • \boldsymbol{x} + \boldsymbol{y} - 12 = 0

Solving this system gives the values of \(x \text{ and } y\) that optimize our objective function. Initially, we find \(位\), then substitute these into the other equations to find all unknowns.

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Most popular questions from this chapter

Given the demand function $$ Q=\frac{P_{\mathrm{A}} Y^{2}}{P} $$ where \(P_{A}=10, Y=2\) and \(P=4\), find the income elasticity of demand. If \(P_{A}\) and \(P\) are fixed, estimate the percentage change in \(Y\) needed to raise \(Q\) by \(2 \%\).

Given the demand function $$ Q=200-2 P-P_{\mathrm{A}}+0.1 Y^{2} $$ where \(P=10, P_{\mathrm{A}}=15\) and \(Y=100\), find (a) the price elasticity of demand (b) the cross-price elasticity of demand (c) the income elasticity of demand Estimate the percentage change in demand if \(P_{\mathrm{A}}\) rises by \(3 \%\). Is the alternative good substitutable or complementary?

Verify that \(x=1, y=-1\) satisfy the equation \(x^{2}-2 y^{3}=3\). Use implicit differentiation to find the value of \(\mathrm{d} y / \mathrm{d} x\) at this point.

A firm's production function is given by $$ Q=2 L^{1 / 2}+3 K^{1 / 2} $$ where \(Q, L\) and \(K\) denote the number of units of output, labour and capital. Labour costs are \(\$ 2\) per unit, capital costs are \(\$ 1\) per unit and output sells at \(\$ 8\) per unit. Show that the profit function is $$ \pi=16 L^{1 / 2}+24 K^{1 / 2}-2 L-K $$ and hence find the maximum profit and the values of \(L\) and \(K\) at which it is achieved.

By substituting $$ Y=\frac{b+I^{*}}{1-a} $$ into $$ C=a Y+b $$ write down the reduced equation for \(C\) in terms of \(a, b\) and \(l^{*}\). Hence show that the investment multiplier for \(C\) is $$ \frac{a}{1-a} $$ Deduce that an increase in investment always leads to an increase in consumption. Calculate the change in consumption when investment rises by 2 units if the marginal propensity to consume is \(1 / 2\).
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