We are tasked with finding the minimum and maximum values of the quadratic function:

$Q(x, y) = 5x^2 + 2y + xy$

subject to the constraint:

$x^2 + y^2 = 1$

This is a problem of constrained optimization, and the method of Lagrange multipliers is well-suited for this. We'll solve the problem step-by-step.

Step 1: Define the Lagrange function

The Lagrange function $\mathcal{L}(x, y, \lambda)$ combines the objective function $Q(x, y)$ and the constraint $g(x, y) = x^2 + y^2 - 1 = 0$:

$\mathcal{L}(x, y, \lambda) = 5x^2 + 2y + xy + \lambda (x^2 + y^2 - 1)$

Here, $\lambda$ is the Lagrange multiplier.

Step 2: Find partial derivatives

Now, we compute the partial derivatives of $\mathcal{L}(x, y, \lambda)$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero to find the critical points.

1. Partial derivative with respect to $x$:

$\frac{\partial \mathcal{L}}{\partial x} = 10x + y + 2\lambda x = 0 \quad \text{(Eq. 1)}$

2. Partial derivative with respect to $y$:

$\frac{\partial \mathcal{L}}{\partial y} = x + 2 + 2\lambda y = 0 \quad \text{(Eq. 2)}$

3. Partial derivative with respect to $\lambda$ (the constraint equation):

$\frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 1 = 0 \quad \text{(Eq. 3)}$

Step 3: Solve the system of equations

We now solve the system formed by the three equations (Eq. 1, Eq. 2, and Eq. 3).

From Eq. 1:

$y = -10x - 2\lambda x$ Substitute this into Eq. 2:

$x + 2 + 2\lambda(-10x - 2\lambda x) = 0$ This gives us a relation between (x, y, \ .