Question

The maximum value of \( f(x, y) = 49 - x^2 - y^2 \) on the line \( x + 3y = 10 \) is __________.

Hint:

To maximize a function subject to a constraint, use the method of Lagrange multipliers by setting up the Lagrangian function and solving the resulting system of equations.

Answer 1

Correct Answer: 39

We are asked to find the maximum value of the function \( f(x, y) = 49 - x^2 - y^2 \) subject to the constraint \( x + 3y = 10 \). To solve this, we use the method of Lagrange multipliers.
Step 1: Set up the Lagrange multiplier equations.
The constraint is \( g(x, y) = x + 3y - 10 = 0 \). The Lagrangian function is given by: \[ \mathcal{L}(x, y, \lambda) = 49 - x^2 - y^2 + \lambda(x + 3y - 10) \] Step 2: Take partial derivatives.
We compute the partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \): \[ \frac{\partial \mathcal{L}}{\partial x} = -2x + \lambda = 0 \quad \text{(1)} \] \[ \frac{\partial \mathcal{L}}{\partial y} = -2y + 3\lambda = 0 \quad \text{(2)} \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = x + 3y - 10 = 0 \quad \text{(3)} \] Step 3: Solve the system of equations.
From equation (1), we have \( \lambda = 2x \). From equation (2), we have \( \lambda = \frac{2y}{3} \). Equating the two expressions for \( \lambda \), we get: \[ 2x = \frac{2y}{3} \quad \Rightarrow \quad x = \frac{y}{3} \] Substitute \( x = \frac{y}{3} \) into the constraint equation (3): \[ \frac{y}{3} + 3y = 10 \quad \Rightarrow \quad \frac{y}{3} + \frac{9y}{3} = 10 \quad \Rightarrow \quad \frac{10y}{3} = 10 \quad \Rightarrow \quad y = 3 \] Substitute \( y = 3 \) into \( x = \frac{y}{3} \), we get: \[ x = \frac{3}{3} = 1 \] Step 4: Calculate the maximum value.
Now, substitute \( x = 1 \) and \( y = 3 \) into the objective function: \[ f(1, 3) = 49 - 1^2 - 3^2 = 49 - 1 - 9 = 39 \] Thus, the maximum value of \( f(x, y) \) on the line \( x + 3y = 10 \) is \( \boxed{39} \).