Problem 27 - Precalculus

Question

Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to find the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables \(x, y\) and \(z\) are the dimensions of the box and \(\lambda\) is called a Lagrange multiplier. With the help of your classmates, solve the system. $$ \left\\{\begin{aligned} 2 y+2 z &=\lambda y z \\ 2 x+2 z &=\lambda x z \\ 2 y+2 x &=\lambda x y \\ x y z &=1000 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The dimensions of the box are \(x = y = z = 10\), and \(\lambda = \frac{2}{5}\).

Step 1: Understand the equations

We are given a system of equations to solve for the dimensions \(x\), \(y\), and \(z\) of a box, with the constraint that the volume \(x y z = 1000\). The variable \(\lambda\) is a Lagrange multiplier that helps us find critical points with constraints. Our goal is to find \(x\), \(y\), \(z\), and \(\lambda\) that satisfy all four equations.

Step 2: Solve the Lagrange multiplier equations

The first three equations incorporate the Lagrange multiplier \(\lambda\). Start by rearranging each equation to express \(\lambda\):1. \(\lambda = \frac{2y + 2z}{yz}\)2. \(\lambda = \frac{2x + 2z}{xz}\)3. \(\lambda = \frac{2y + 2x}{xy}\).

Step 3: Equate and simplify equations

Since \(\lambda\) is equal in each rearranged equation, we equate them:- From Equation 1 and 2: \(\frac{2y + 2z}{yz} = \frac{2x + 2z}{xz}\)- From Equation 2 and 3: \(\frac{2x + 2z}{xz} = \frac{2y + 2x}{xy}\).Simplify these equations to find relationships between \(x\), \(y\), and \(z\).

Step 4: Simplify and find relationships

Simplifying the equations, we get:- \(y = x\)- \(z = x\).This suggests that the box is a cube since all dimensions must be equal.

Step 5: Substitute into the volume equation

With \(x = y = z\), substitute into the volume equation \(x yz = 1000\):- \(x^3 = 1000\).Solve this equation for \(x\).

Step 6: Solve for the dimensions

To find \(x\), take the cube root of both sides:- \(x = 10\).Hence, the dimensions \(x = y = z = 10\).

Step 7: Calculate \(\lambda\)

Plug \(x = y = z = 10\) back into any expression for \(\lambda\), for instance:- \(\lambda = \frac{2 \times 10 + 2 \times 10}{10 \times 10} = \frac{40}{100} = \frac{2}{5}\).

Key Concepts

Systems of Nonlinear Equations

When dealing with systems of nonlinear equations, we often encounter multiple equations that need to be solved simultaneously. Nonlinear equations, unlike linear ones, do not graph as straight lines. Instead, they might represent curves or other shapes. This makes finding solutions, especially when multiple equations are involved, more complex.
In our exercise, we have a set of four equations: three involving the Lagrange multiplier \(\lambda\), and one constraint equation for the volume \(xyz = 1000\). The nonlinear nature arises primarily from the multiplicative combination of variables that does not simplify to a linear form. Solving these requires careful manipulation and often the use of numerical methods or symbolic algebra.
Key strategies include expressing some variables in terms of others and substituting back to simplify the system until a solution for each variable emerges. For example, equating expressions for \(\lambda\) as done in the exercise helps to progressively reduce complexity until a straightforward solution appears.

Constrained Optimization

Constrained optimization is a powerful tool in calculus for finding the extrema of a function subject to certain constraints. In simple terms, it allows us to determine the optimal solution within a set boundary.
For this exercise, our task was to minimize the surface area of a rectangular box while maintaining a fixed volume of 1000 cubic inches. Here, the Lagrange multipliers method becomes essential. By introducing a Lagrange multiplier \(\lambda\), we combine the constraint with our main function and then solve the resulting system of equations.
The idea is to convert a constrained problem into an unconstrained one by integrating the constraint as an additional term with the multiplier. When we find points where the gradient of the objective function aligns with the gradient of the constraint, we find possible solutions that respect our constraints.

Third Semester Calculus

Third semester calculus, also known as multivariable calculus, builds upon the foundation laid by calculus I and II, focusing on functions of multiple variables. It explores several complex but intriguing topics such as partial derivatives, gradients, and constrained optimization.
This course often involves understanding the behavior of functions in three-dimensional space, which is particularly handy in fields like engineering and physics. The exercise we tackled is a classic example, showing how to apply these concepts to determine the dimensions of a 3D object optimally.
By using tools like Lagrange multipliers, students learn not only to handle multiple variable scenarios but also to apply calculus in real-world contexts where optimal solutions are necessary, whether minimizing cost, energy, or surface area.

Critical Points

Critical points in calculus refer to points in the domain of a function where the derivative is zero or undefined. These are significant because they signal potential locations of minima, maxima, or saddle points.
In constrained optimization problems, like the one in our exercise, finding critical points helps determine where the function reaches its extreme values, subject to the given constraints. The role of the Lagrange multiplier method is to find these critical points effectively even when constraints are present.
Once we apply this method and solve for \(\lambda\) as well as the dimensions, we check these points by observing how the function behaves around them. In our case, the critical point discovered revealed that the dimensions of the box that satisfied both the constraint and the optimal condition were \(x = y = z = 10\), indicating a cube with minimized surface area for the given volume.