Question

Let f ∶ \(\R^2 → \R\) be the function defined by f(x, y) = x² - 12y. If M and m be the maximum value and the minimum value, respectively, of the function f on the circle x² + y² = 49, then |M| + |m| equals __________

Answer 1

Correct Answer: 169

To solve the problem of finding the maximum and minimum values of the function \(f(x, y) = x^2 - 12y\) on the circle \(x^2 + y^2 = 49\), we will use the method of Lagrange multipliers. The constraint is the circle \(g(x, y) = x^2 + y^2 - 49 = 0\). We define the Lagrangian as \(L(x, y, \lambda) = x^2 - 12y + \lambda(x^2 + y^2 - 49)\). To find the extrema, calculate the gradients and set them equal to zero: \(\nabla L = (2x + 2\lambda x, -12 + 2\lambda y, x^2 + y^2 - 49)\).

1. Set \(\frac{\partial L}{\partial x} = 2x + 2\lambda x = 0 \Rightarrow x(1 + \lambda) = 0\).

2. Set \(\frac{\partial L}{\partial y} = -12 + 2\lambda y = 0 \Rightarrow \lambda y = 6\).

3. Set \(\frac{\partial L}{\partial \lambda} = x^2 + y^2 - 49 = 0\).

Now, solve the system:

**Case 1**: If \(x = 0\), then \(y^2 = 49 \Rightarrow y = \pm7\).

- For \(y = 7\), \(f(0, 7) = 0^2 - 12(7) = -84\).

- For \(y = -7\), \(f(0, -7) = 0^2 - 12(-7) = 84\).

**Case 2**: If \(1 + \lambda = 0\), then \(\lambda = -1\), substitute into \(\lambda y = 6\) yields \(y = -6\).

Insert in \(x^2 + (-6)^2 = 49 \Rightarrow x^2 = 13 \Rightarrow x = \pm\sqrt{13}\).

- For \(x = \sqrt{13}\), \(f(\sqrt{13}, -6) = 13 - 12(-6) = 85\).

- For \(x = -\sqrt{13}\), \(f(-\sqrt{13}, -6) = 13 - 12(-6) = 85\).

Hence, the maximum value \(M = 85\) and the minimum value \(m = -84\).

Therefore, \(|M| + |m| = |85| + |-84| = 85 + 84 = 169\).