Final answer:
To find the minimum of the function f(x, y) = 3x^2 + xy + 2y^2 - 15x - 14y - 5, we need to find the critical points by taking the partial derivatives with respect to x and y. Setting the derivatives equal to 0 and solving the system of equations, we find x = 3 and y = 2 as the values at the minimum.
Step-by-step explanation:
The function f(x, y) = 3x^2 + xy + 2y^2 - 15x - 14y - 5 has a minimum. To find the minimum, we need to find the critical points of the function. We can do this by taking the partial derivatives of the function with respect to x and y, setting them equal to 0, and solving for x and y.
First, let's find the partial derivative with respect to x: f'(x) = 6x + y - 15. Setting this equal to 0 and solving for x, we get: 6x + y = 15.
Next, let's find the partial derivative with respect to y: f'(y) = x + 4y - 14. Setting this equal to 0 and solving for y, we get: x + 4y = 14.
Now, we have a system of two equations:
6x + y = 15
x + 4y = 14
Solving this system of equations, we find x = 3 and y = 2. Therefore, the values of x and y at the minimum of the function are x = 3 and y = 2.