Problem 74 - Precalculus with Limits

Question

Advanced Applications In Exercises 73 and \(74 ,\) find values of \(x , y ,\) and \(\lambda\) that satisfy the system. These systems arise in certain optimization problems in calculus, and \(\lambda\) is called a Lagrange multiplier. $$ \left\\{ \begin{aligned} 2 + 2 y + 2 \lambda & = 0 \\ 2 x + 1 + \lambda & = 0 \\\ 2 x + y - 100 & = 0 \end{aligned} \right. $$

Step-by-Step Solution

Answer

The solutions for the variables are: \(x = 25\), \(y = 50\) and \(\lambda = -51\).

Step 1: Simplify the Equations

First, simplify the given equations. The first equation will be \(2y + 2\lambda = -2\). The second equation will be \(2x + \lambda = -1\). The third equation will remain the same: \(2x + y = 100\).

Step 2: Solve the First and Second Equation for \(\lambda\)

To make the substitution easier, solve the first and second equations for \(\lambda\). From the first equation, \(\lambda = -1 - y\). Similarly, from the second equation, \(\lambda = -1 - 2x\).

Step 3: Setting the two \(\lambda\) equations equal

As both equations are \(\lambda = ...\), it's possible to set both equations equal to each other, getting \(-1 - y = -1 - 2x\).

Step 4: Solve for x

Isolate \(x\) in the equation \(-1 - y = -1 - 2x\), by solving for \(x\). In doing this, \(x = y/2\).

Step 5: Substitution of \(x\) into the third equation

The obtained value of \(x\) should be substituted into the third equation: in the third equation, replace \(x\) with \(y/2\). Doing this yields \(y + y = 100\), which simplifies to \(2y = 100\).

Step 6: Solve for y

Solve this equation for \(y\) to get \(y = 50\).

Step 7: Substitution of \(y\) and \(x\) into any equation

The previously obtained \(x\) should be substituted into any of the \(\lambda\) equations. Substituting into the equation \(\lambda = -1 - y\), gives: \(\lambda = -1 - 50 = -51\).

Step 8: Checking the solutions in all equations

Lastly, check if the obtained values are correct by substituting \(x\), \(y\), and \(\lambda\) into the equations in the original system.

Key Concepts

System of Equations

When facing an optimization problem, you often have to deal with a system of equations, which is essentially a set of two or more equations involving the same set of variables. The goal here is to find a common solution — that is, values for the variables that make all the equations true simultaneously.

For example, in the given exercise, we have three equations with variables x, y, and \( \lambda \), where \( \lambda \) is the Lagrange multiplier. To solve this system, one can employ various methods like graphing, substitution, elimination, or more advanced techniques for larger systems such as matrix operations. It's critical to remember that the solution to a system of equations represents the set of all points that satisfy all equations at once; geometrically, it can be thought of as the intersection point(s) of graphs represented by these equations.

Calculus Optimization Problems

Optimization problems in calculus are about finding the maximum or minimum values of a function within a given set of constraints. These constraints often come in the form of equations or inequalities.

Calculus optimization problems can be solved using methods such as the First Derivative Test or the Second Derivative Test, which involve taking the derivative of a function to find critical points. However, when constraints are involved, Lagrange multipliers become a powerful tool. This method involves introducing an auxiliary multiplier (usually denoted as \( \lambda \)) for each constraint and forming a new function that incorporates both the original function to be optimized and the constraints. The solutions satisfying the Lagrange equations hold potential as maximum or minimum points under the given constraints.

Substitution Method

The substitution method is a technique used to solve systems of equations where one equation is manipulated to isolate one variable, which is then 'substituted' into the other equations.

Using this method simplifies the system, turning it into a more manageable set of equations with one less variable each time a substitution is made. This technique can be particularly helpful when dealing with a system of nonlinear equations or when the equations are not easily aligned for elimination.

Applying the substitution method as seen in the given exercise, once we solve for \( \lambda \) in two different equations, we equate these expressions to isolate \( x \) and \( y \) in subsequent steps. It's a sequential process where the deduction of one variable leads to the discovery of others until all values are found.