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Chapter 8: Problem 7

a. Write the Lagrange system of partial derivative equations. b. Locate the optimal point of the constrained system. c. Identify the optimal point as either a maximum point or a minimum point. $$ \left\\{\begin{array}{l} \text { optimize } f(x, y)=80 x+5 y^{2} \\ \text { subject to } g(x, y)=2 x+2 y=1.4 \end{array}\right. $$

Short Answer

Expert verified
The optimal point is \((x, y) = (-7.3, 8)\). It is likely a maximum point under the constraint.

Step by step solution

01

Define the Lagrangian

To solve the given constrained optimization problem, we first define the Lagrangian. The Lagrangian is a function that incorporates the objective function and the constraint using a Lagrange multiplier \( \lambda \). For the given problem: \[ \mathcal{L}(x, y, \lambda) = 80x + 5y^2 + \lambda (1.4 - 2x - 2y). \]
02

Find Partial Derivatives

Next, we find the partial derivatives of the Lagrangian with respect to \( x \), \( y \), and \( \lambda \). These are used to form the system of equations needed for optimization. 1. \( \frac{\partial \mathcal{L}}{\partial x} = 80 - 2\lambda = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 10y - 2\lambda = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1.4 - 2x - 2y = 0 \)
03

Solve the System of Equations

Solve the equations obtained from the partial derivatives:1. From \( 80 = 2\lambda \), we have \( \lambda = 40 \).2. Substitute \( \lambda = 40 \) into \( 10y = 2\lambda \), we have \( 10y = 80 \), giving \( y = 8 \).3. Substitute \( y = 8 \) into the constraint equation \( 2x + 2y = 1.4 \), giving \( 2x + 16 = 1.4 \), leading to \( 2x = -14.6 \), so \( x = -7.3 \).
04

Verify Constraints

Verify if our solution satisfies the original constraint. Plugging \( x = -7.3 \) and \( y = 8 \) back into the constraint gives:\( 2(-7.3) + 2(8) = -14.6 + 16 = 1.4 \), which confirms that the constraint is satisfied.
05

Identify Optimality Condition

To identify if the point \((x, y) = (-7.3, 8)\) is a maximum or minimum, check the properties of the function. Since the objective function \( f(x, y) = 80x + 5y^2 \) is linear in \( x \) and quadratic in \( y \), it indicates an unbounded behavior. However, given the constraint, the presence of the linear combination implies it aims to maximize \( 80x \). Typically, computational methods or second-order conditions would verify this further.

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Key Concepts

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

Constrained Optimization
Constrained optimization is a fascinating process where we try to find the best solution to a problem with given restrictions. Imagine trying to fit a certain amount of cargo into a limited space in a truck. You want to maximize what you carry without exceeding the truck's capacity. This is similar to what constrained optimization seeks to do in mathematics. In our exercise, we have a function we want to optimize:
  • 80x - represents a linear desire to increase 'x'.
  • 5y虏 - introduces a non-linear component for 'y'.
However, we're not free to fiddle with 'x' and 'y' as we please. We have a constraint: \(2x + 2y = 1.4\), which acts like a rule to follow. By introducing a new variable called the 'Lagrange multiplier', we incorporate this constraint into our optimization problem. This strategy links together the main objective and the restriction, making it feasible to solve through a unified approach. Lagrange multipliers are like the magic glue in constrained optimization.
Partial Derivatives
In the realm of mathematics, partial derivatives are a key tool. They help us understand how a function changes as each variable changes independently. Think about adjusting the temperature and brightness in a room. Changing one doesn鈥檛 affect the other directly, but both influence the overall ambiance. Similarly, we calculate partial derivatives to see the isolated effect of each variable. In our exercise, we computed the partial derivatives of the Lagrangian function with respect to each variable. Here's how it panned out:
  • For \(x\), the derivative \(\frac{\partial \mathcal{L}}{\partial x} = 80 - 2\lambda\) indicates sensitivity to \(x\).
  • For \(y\), \(\frac{\partial \mathcal{L}}{\partial y} = 10y - 2\lambda\) shows \(y\)'s effect.
  • \(\lambda\), the constraint, ensures \(1.4 - 2x - 2y = 0\) holds true.
These derivatives create equations forming a system. Solving them gives us the optimal answers.
Optimal Point Identification
Finding and verifying the optimal point is the final step in constrained optimization. Once we have solved for the values of \(x\) and \(y\) using the system of equations, we need to determine if this point is optimal. The optimal point is not just any solution; it鈥檚 a point where the function satisfies the constraint while achieving the desired extremum (maximum or minimum). Let's summarize our findings from the exercise:
  • The solution \((-7.3, 8)\) satisfies the constraint \(2x + 2y = 1.4\).
  • The function \(f(x, y) = 80x + 5y^2\) suggests unbounded behavior due to its linear term.
In practical terms, this indicates the direction in which the solution likely aims for maximization because increasing \(x\) will increase the function's value. But being constrained, it reaches the optimal point allowed by the condition. Computational methods can further validate this finding, ensuring it aligns with our expectations. Understanding this point is vital for deciding if further action towards another solution is required.

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Most popular questions from this chapter

Single Young Adults The percentage of adults 20 to 24 years of age who had not yet been married is given in the table. Percentage of 20 - to 24 -Year-OId Adults Who Have Not Yet Been Married $$ \begin{array}{|c|c|c|c|c|c|} \hline \text { Year } & 1960 & 1970 & 1980 & 1990 & 2000 \\ \hline \text { Not Yet Married (\%) } & 41 & 46 & 60 & 72 & 78 \\ \hline \end{array} $$a. Use the method of least squares to find the multivariable function \(f\) with inputs \(a\) and \(b\) for the bestfitting line \(y=a x+b,\) where \(x\) is the number of years since \(1970 .\) b. Write the minimum value of \(f(a, b)\). c. Write the linear model that best fits these data. d. Use the model to estimate the percentage of 20 to 24 year old adults 20 - to 24 -year-old who had never been married in \(2010 .\)

a. Write the Lagrange system of partial derivative equations. b. Locate the optimal point of the constrained system. c. Identify the optimal point as either a maximum point or a minimum point. $$ \left\\{\begin{array}{l} \text { optimize } f(x, y)=x^{2} y \\ \text { subject to } g(x, y)=x+y=16 \end{array}\right. $$

New York Precipitation (Historic) The table gives the number of inches of precipitation that fell in New York City in early 1855 Precipitation in New York City 1855 $$ \begin{array}{|l|c|c|c|} \hline \text { Month } & \text { Jan } & \text { Feb } & \text { Mar } \\ \hline \text { Precipitation (inches) } & 5.50 & 3.25 & 1.35 \\ \hline \end{array} $$ a. Use the method of least squares to write the multivariable function \(f\) with inputs \(a\) and \(b\) for the best-fitting line \(y=a x+b\), where \(x\) is 1 in January, 2 in February, and 3 in March. b. Calculate the minimum value of \(f(a, b)\) and verify that it is a minimum. c. Write the linear model that best fits these data.

Locate and classify any critical points. $$ H(r, s)=r s+2 s^{2}+r^{2} $$

Radio Production The daily output at a plant manufacturing transistor radios can be modeled as $$ f(L, K)=10.5463 L^{0.3} K^{0.5} \text { radios } $$ where \(L\) is the size of the labor force measured in hundred worker hours and \(K\) is the capital investment in thousand dollars. a. Suppose that the plant manager has a daily budget of \(\$ 15,000\) to be invested in capital or spent on labor and that the average wage of an employee at the radio plant is \(\$ 7.50\) per hour. What combination of worker hours and capital expenditures will yield maximum daily production? What is the maximum production level? b. Use two close points to show that the output value in part \(a\) is a maximum. c. Calculate and interpret the marginal productivity of money for this manufacturing process.
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