Solution: We need to find the maximum of the function $49 - x^2 - y^2$ such that $ x + 3y = 10$. We will solve this problem by the Lagrange's multipliers. Let us denote \[ g(x,y) = x + 3y - 10. \] Then the Lagrange multiplier will be \begin{align}\label{eq:18May2024-1} L(x,y, \lambda) & = f(x,y) + \lambda g(x,y) \notag \\ & = 49 - x^2 - y^2 +\lambda (x + 3y - 10). \end{align} Then the extrema can be found by solving \begin{align*} \frac{\partial L}{\partial x} = 0 = \frac{\partial L}{\partial y} = \frac{\partial L}{\partial \lambda }. \end{align*} That is, \begin{align*} \frac{\partial L}{\partial x } = 0 & \implies -2x + \lambda = 0 \implies x = \frac{\lambda}{2} \\[2ex] \frac{\partial L}{\partial y } = 0 & \implies -2y + 3\lambda = 0 \implies y = \frac{3\lambda }{2} \\[2ex] \frac{\partial L}{\partial \lambda } = 0 & \implies x + 3y = 10. \end{align*}

Thus, we have \begin{align*} \frac{\lambda}{2} + \frac{9\lambda}{2} = 10 \implies \lambda = 2. \end{align*} Therefore, \[ x = 1, \quad \text{and} \quad y = 3. \] Therefore, the point of extrema is $(1,3)$ and corresponding maximum value will be \[ f(1,3) = 39. \]