Determine the stationary points of \(f(x, y)=\) \(2 x^{2}+3 y^{2}+5 x+12 y+19\)

Short Answer

Expert verified
Answer: The stationary point of the given function is at \((-\frac{5}{4}, -2)\).

Step by step solution

01

Calculate the partial derivatives with respect to x and y

To find the partial derivative of f with respect to x, differentiate the function with respect to x while treating y as a constant: \(\frac{\partial f}{\partial x} = 4x + 5\). Similarly, to find the partial derivative of f with respect to y, differentiate the function with respect to y while treating x as a constant: \(\frac{\partial f}{\partial y} = 6y + 12\).
02

Set the partial derivatives equal to zero

To locate the stationary points, we set both partial derivatives equal to zero: \(4x + 5 = 0\) and \(6y + 12 = 0\).
03

Solve the system of equations

To solve the equations, first solve each one for the variable involved: For the equation regarding x: \(4x + 5 = 0\), subtract 5 from both sides and then divide by 4 \(x = -\frac{5}{4}\). For the equation regarding y: \(6y + 12 = 0\), subtract 12 from both sides and then divide by 6 \(y = -2\). The stationary point of \(f(x, y) = 2x^2 + 3y^2 + 5x + 12y + 19\) is at the point \((-\frac{5}{4}, -2)\).

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