Question

Finding extrema. Find the point $(x^{\ast },y^{\ast },z^{\ast })$ that is at the minimum of the function $$f(x,y,z)=2x^2+8y^2+z^2$$ subject to the constraint equation $$g(x,y,z)=6x+4y+4z-72=0$$

Short answer

(6, 1, 8)

Step-by-step solution

Introduce Lagrange Multipliers

To find the minimum of the function subject to the constraint, use the method of Lagrange multipliers. Define the Lagrangian function as $$L(x,y,z,\lambda )=f(x,y,z)-\lambda (g(x,y,z))$$ where \lambda is the Lagrange multiplier.

Define Lagrangian

Substitute the given functions into the Lagrangian: $$L(x,y,z,\lambda )=2x^2+8y^2+z^2-\lambda (6x+4y+4z-72)$$

Take Partial Derivatives

Calculate the partial derivatives of the Lagrangian with respect to each variable and set them to zero: $$\frac{\partial L}{\partial x}=4x-6\lambda =0\Rightarrow x=\frac{3\lambda }{2}$$ $$\frac{\partial L}{\partial y}=16y-4\lambda =0\Rightarrow y=\frac{\lambda }{4}$$ $$\frac{\partial L}{\partial z}=2z-4\lambda =0\Rightarrow z=2\lambda$$ $$\frac{\partial L}{\partial \lambda }=-(6x+4y+4z-72)=0$$

Substitute Variables into Constraint

Substitute the values of x, y, and z found from the partial derivatives into the constraint equation: $$6(\frac{3\lambda }{2})+4(\frac{\lambda }{4})+4(2\lambda )=72$$ Simplify and solve for \lambda: $$9\lambda +\lambda +8\lambda =72\Rightarrow 18\lambda =72\Rightarrow \lambda =4$$

Find x, y, and z

Substitute \lambda back into the expressions for x, y, and z: $$x=\frac{3(4)}{2}=6$$ $$y=\frac{4}{4}=1$$ $$z=2(4)=8$$

Verify Solution

Check the solution (6, 1, 8) against the constraint: $$6(6)+4(1)+4(8)=36+4+32=72$$ The solution satisfies the constraint.

Key concepts

Extrema

In calculus, finding the extrema of a function involves determining its maximum and minimum values. These values are crucial because they help us understand the behavior of the function. Extrema are found by analyzing the function's critical points where its first derivative is zero or undefined. For multivariable functions, we look for points where all partial derivatives are zero.

When dealing with constrained optimization problems, like the one in the exercise, we can't apply this method directly. This is where Lagrange multipliers come in handy. By introducing a new variable (the Lagrange multiplier), we can convert the problem into finding the extrema of a modified function.

Constraint Optimization

Constraint optimization deals with finding the extrema of a function subject to certain conditions. For example, you might want to maximize or minimize a function while adhering to specific constraints.

In our exercise, we want to find the minimum of the function $f(x,y,z)=2x^2+8y^2+z^2$, subject to the constraint $g(x,y,z)=6x+4y+4z-72=0$. This process involves:

• Defining a new function (Lagrangian) that combines our original function with the constraint.
• Introducing a Lagrange multiplier to adjust the influence of the constraint.
• Taking partial derivatives of this Lagrangian and setting them to zero.
• Solving the resulting system of equations to find the critical points which are potential candidates for extrema.

This approach ensures that we find all possible extrema that satisfy our constraints.

Partial Derivatives

Partial derivatives are vital in multivariable calculus because they allow us to understand how functions change with respect to one of their variables while keeping the others constant. For a function $f(x,y,z)$, the partial derivative with respect to $x$ is denoted by $\frac{\partial f}{\partial x}$.

In the context of Lagrange multipliers, we take partial derivatives of the Lagrangian function with respect to each variable as well as the Lagrange multiplier. This allows us to find critical points by setting these derivatives to zero.

• For $x$: $\frac{\partial L}{\partial x}=4x-6\lambda =0$ leads to $x=\frac{3\lambda }{2}$.
• For $y$: $\frac{\partial L}{\partial y}=16y-4\lambda =0$ leads to $y=\frac{\lambda }{4}$.
• For $z$: $\frac{\partial L}{\partial z}=2z-4\lambda =0$ leads to $z=2\lambda$.
• For $\lambda$: $\frac{\partial L}{\partial \lambda }=-(6x+4y+4z-72)=0$.

By solving these equations, we identify the values of $x,y,z$, and $\lambda$ that satisfy both the function and the constraint.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of several variables. It deals with functions like $f(x,y,z)$ which depend on more than one variable. This branch of calculus includes studying partial derivatives, multiple integrals, and differential equations in higher dimensions.

In this field, we frequently use partial derivatives to study how these functions change with respect to each variable independently. For problems involving constraints, multivariable calculus provides techniques like Lagrange multipliers to identify optimal solutions.

• Example in the exercise: We analyzed a function and a constraint equation involving three variables.
• Applied Lagrange multipliers to simplify solving for critical points.

This approach enables us to tackle complex real-world problems, such as optimizing production levels or minimizing costs under various constraints.