Find the maximum and minimum values of 𝑄ሺ𝑥⃗ሻ ൌ 3𝑥ଵଶ ൅ 2𝑥ଵ 𝑥ଶ ൅ 3𝑥ଶଶ subject to thecondition of ‖𝑥⃗‖ ൌ 2. (10 marks)

The problem is to find the maximum and minimum values of the function Q(x) = 3x₁² + 2x₁x₂ + 3x₂² subject to the condition ||x|| = 2.

This is a problem of optimization on a circle of radius 2 in the x₁x₂-plane.

Step 1: Rewrite the condition The condition ||x|| = 2 can be rewritten as x₁² + x₂² = 4.

Step 2: Substitute x₂ from the condition into the function From the condition, we can express x₂ in terms of x₁: x₂ = ±sqrt(4 - x₁²). Substituting this into the function gives us two functions of one variable: Q₁(x₁) = 3x₁² + 2x₁sqrt(4 - x₁²) + 3(4 - x₁²) and Q₂(x₁) = 3x₁² - 2x₁sqrt(4 - x₁²) + 3(4 - x₁²).

Step 3: Find the critical points of the functions The critical points of a function are the points where its derivative is zero or undefined. We find the derivatives of Q₁ and Q₂ and set them equal to zero, solving for x₁.

Step 4: Evaluate the functions at the critical points and endpoints The possible maximum and minimum values of the function occur at the critical points and the endpoints of the interval. We evaluate Q₁ and Q₂ at these points.

Step 5: Compare the function values The largest of the function values is the maximum value of the function, and the smallest is the minimum value.

Note: The actual computation of the derivatives and function values depends on your knowledge of calculus.