This extreme value Problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. $f(x,y,z) = 4x + 4y + 7z$, $2x^2 + 2y^2 + 7z^2 = 23$

Neetesh Kumar | November 30, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.8 Question Number 4
Contact me if you need help with Homework, Assignments, Tutoring Sessions, or Exams for STEM subjects.
You can see our Testimonials or Vouches from here of the previous works I have done.

Get Homework Help

---

Step-by-step solution:

We are tasked with finding the extreme values of the function $f(x, y, z) = 4x + 4y + 7z$ subject to the constraint $2x^2 + 2y^2 + 7z^2 = 23$.

To solve this, we will use the method of Lagrange multipliers.

Step 1: Define the Lagrange Multiplier Equations

The method of Lagrange multipliers involves solving the system of equations:

$\nabla f(x, y, z) = \lambda \nabla g(x, y, z)$

where $\nabla f(x, y, z)$ and $\nabla g(x, y, z)$ are the gradients of $f(x, y, z)$ and $g(x, y, z)$, respectively, and $\lambda$ is the Lagrange multiplier.

Compute the gradients:

• The gradient of $f(x, y, z) = 4x + 4y + 7z$ is:

$\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (4, 4, 7)$

• The gradient of $g(x, y, z) = 2x^2 + 2y^2 + 7z^2 - 23$ is:

$\nabla g(x, y, z) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) = (4x, 4y, 14z)$

Step 2: Set up the System of Equations

From the Lagrange multiplier condition $\nabla f(x, y, z) = \lambda \nabla g(x, y, z)$, we have the system of equations:

1. $4 = \lambda \cdot 4x$
2. $4 = \lambda \cdot 4y$
3. $7 = \lambda \cdot 14z$

Additionally, we need to satisfy the constraint:

$2x^2 + 2y^2 + 7z^2 = 23$

Step 3: Solve the System of Equations

From the first equation, we solve for $\lambda$:

$4 = \lambda \cdot 4x \quad \Rightarrow \quad \lambda = \frac{1}{x}$

From the second equation, we solve for $\lambda$:

$4 = \lambda \cdot 4y \quad \Rightarrow \quad \lambda = \frac{1}{y}$

From the third equation, we solve for $\lambda$:

$7 = \lambda \cdot 14z \quad \Rightarrow \quad \lambda = \frac{1}{2z}$

Step 4: Relate the Variables

Now that we have expressions for $\lambda$, we can equate them to find relationships between $x$, $y$, and $z$.

From $\lambda = \frac{1}{x}$ and $\lambda = \frac{1}{y}$, we get:

$\frac{1}{x} = \frac{1}{y} \quad \Rightarrow \quad x = y$

From $\lambda = \frac{1}{x}$ and $\lambda = \frac{1}{2z}$, we get:

$\frac{1}{x} = \frac{1}{2z} \quad \Rightarrow \quad x = 2z$

Thus, we have the relationships:

$x = y \quad \text{and} \quad x = 2z$

Step 5: Substitute into the Constraint

Now substitute $x = y$ and $x = 2z$ into the constraint $2x^2 + 2y^2 + 7z^2 = 23$:

$2x^2 + 2x^2 + 7z^2 = 23$

Since $x = 2z$, substitute $x = 2z$ into the equation:

$2(2z)^2 + 2(2z)^2 + 7z^2 = 23$

Simplify:

$2(4z^2) + 2(4z^2) + 7z^2 = 23$

$8z^2 + 8z^2 + 7z^2 = 23$

$23z^2 = 23$

$z^2 = 1$

Thus, $z = 1$ or $z = -1$.

Step 6: Find Corresponding Values of $x$ and $y$

Using $x = 2z$ and $x = y$, we find:

• If $z = 1$, then $x = 2(1) = 2$ and $y = 2$.
• If $z = -1$, then $x = 2(-1) = -2$ and $y = -2$.

Thus, the possible points are $(x, y, z) = (2, 2, 1)$ and $(x, y, z) = (-2, -2, -1)$.

Step 7: Evaluate $f(x, y, z)$ at the Critical Points

Now, evaluate the function $f(x, y, z) = 4x + 4y + 7z$ at the points $(2, 2, 1)$ and $(-2, -2, -1)$:

• At $(2, 2, 1)$:

$f(2, 2, 1) = 4(2) + 4(2) + 7(1) = 8 + 8 + 7 = 23$

• At $(-2, -2, -1)$:

$f(-2, -2, -1) = 4(-2) + 4(-2) + 7(-1) = -8 - 8 - 7 = -23$

Final Answer:

The maximum value of $f(x, y, z)$ is $\boxed{23}$, and the minimum value is $\boxed{-23}$

Please comment below if you find any error in this solution.
If this solution helps, then please share this with your friends.
Please subscribe to my Youtube channel for video solutions to similar questions.
Keep Smiling :-)