We need to find the maximum and minimum values of the function:

$f(x, y, z) = yz + xy$

subject to the constraints:

$y^2 + z^2 = 225 \quad \text{(1)}$ $xy = 7 \quad \text{(2)}$

To solve this problem, we will apply the method of Lagrange multipliers, which involves introducing multipliers $\lambda$ and $\mu$ for the two constraints.

Step 1: Define the Lagrange function

The Lagrange function $\mathcal{L}(x, y, z, \lambda, \mu)$ is given by:

$\mathcal{L}(x, y, z, \lambda, \mu) = f(x, y, z) - \lambda(g_1(y, z) - 225) - \mu(g_2(x, y) - 7)$

where:

• $f(x, y, z) = yz + xy$
• $g_1(y, z) = y^2 + z^2$
• $g_2(x, y) = xy$

So the Lagrange function becomes:

$\mathcal{L}(x, y, z, \lambda, \mu) = yz + xy - \lambda(y^2 + z^2 - 225) - \mu(xy - 7)$

Step 2: Take partial derivatives and set them equal to zero

We need to take the partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, $z$, $\lambda$, and $\mu$, and set each to zero.

Partial derivatives:

$\frac{\partial \mathcal{L}}{\partial x} = y - \mu y = 0 \quad \Rightarrow \quad y(1 - \mu) = 0$ $\frac{\partial \mathcal{L}}{\partial y} = z + x - 2\lambda y - \mu x = 0$ $\frac{\partial \mathcal{L}}{\partial z} = y - 2\lambda z = 0$ $\frac{\partial \mathcal{L}}{\partial \lambda} = -(y^2 + z^2 - 225) = 0 \quad \Rightarrow \quad y^2 + z^2 = 225$ $\frac{\partial \mathcal{L}}{\partial \mu} = -(xy - 7) = 0 \quad \Rightarrow \quad xy = 7$

Step 3: Solve the system of equations

From $\frac{\partial \mathcal{L}}{\partial x} = 0$, we have $y(1 - \mu) = 0$. This gives two cases:

Case 1: $y = 0$

• If $y = 0$, then from the constraint $xy = 7$, we get $x = \frac{7}{y}$, but this is undefined, so $y = 0$ leads to an invalid solution.

Case 2: $\mu = 1$

• From the equation $y(1 - \mu) = 0$, we get $\mu = 1$.