Question

If the function \( f(x, y) = x^2 + xy + y^2 + \frac{1}{x} + \frac{1}{y} \), \( x \neq 0, y \neq 0 \) attains its local minimum value at the point \( (a, b) \), then the value of \( a^3 + b^3 \) is __________ (round off to TWO decimal places).

Hint:

To find the local minimum of a function of two variables, compute the partial derivatives, set them equal to zero, and solve the system of equations.

Answer 1

Correct Answer: 0.65

We are given the function \[ f(x, y) = x^2 + xy + y^2 + \frac{1}{x} + \frac{1}{y} \] and we need to find the local minimum value of \( f(x, y) \) at the point \( (a, b) \). To find the critical points, we first compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is: \[ f_x(x, y) = 2x + y - \frac{1}{x^2} \] The partial derivative with respect to \( y \) is: \[ f_y(x, y) = 2y + x - \frac{1}{y^2} \] Now, set both partial derivatives equal to zero to find the critical points: \[ 2x + y - \frac{1}{x^2} = 0 \quad \text{(1)} \] \[ 2y + x - \frac{1}{y^2} = 0 \quad \text{(2)} \] By solving the system of equations (1) and (2), we get the critical points. After solving, we find that \( x = 1 \) and \( y = 1 \) satisfy both equations. Therefore, the critical point is \( (a, b) = (1, 1) \). Next, we compute \( a^3 + b^3 \) at this point: \[ a^3 + b^3 = 1^3 + 1^3 = 1 + 1 = 2 \] Thus, the value of \( a^3 + b^3 \) is \(\boxed{2.00}\).