Question

find all points where the function has any relative extrema. identify any saddle points. f(x,y)=2x² + 3xy + 4y² - 6x + 7y

Explanation:

Step1: Find first - order partial derivatives

The function is \(f(x,y)=2x^{2}+3xy + 4y^{2}-6x + 7y\).
The first - order partial derivatives are:
\(f_{x}=\frac{\partial f}{\partial x}=4x + 3y-6\)
\(f_{y}=\frac{\partial f}{\partial y}=3x + 8y+7\)

Step2: Solve the system of equations \(f_{x}=0\) and \(f_{y}=0\)

We have the system of equations:
\(

$$\begin{cases}4x + 3y-6=0\\3x + 8y+7 = 0\end{cases}$$

\)
From the first equation \(4x+3y=6\), we get \(x=\frac{6 - 3y}{4}\).
Substitute \(x=\frac{6 - 3y}{4}\) into the second equation:

$$
LATEXBLOCK1
$$

Substitute \(y = - 2\) into \(x=\frac{6 - 3y}{4}\), we have \(x=\frac{6-3\times(-2)}{4}=\frac{6 + 6}{4}=3\)

Step3: Find second - order partial derivatives

\(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=4\), \(f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=3\), \(f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=8\)

Step4: Use the second - derivative test

The discriminant \(D=f_{xx}f_{yy}-(f_{xy})^{2}\)
Substitute \(f_{xx} = 4\), \(f_{xy}=3\), \(f_{yy}=8\) into the formula:
\(D=(4\times8)-3^{2}=32 - 9=23>0\)
And \(f_{xx}=4>0\)

Answer:

A. The point(s) is(are) \((3,-2)\)
B. There are no relative maxima.
A. The point(s) is(are) \((3,-2)\)