Question

find all points where the function has any relative extrema. identify any saddle points.
f(x,y)=2x^{2}+3xy + 4y^{2}-6x + 7y
find all points where the function has relative maxima. select the correct choice below and fill in any answer boxes within your choice.
o a. the point(s) is(are)
(simplify your answer. type an ordered pair. use a comma to separate answers as needed.)
o b. there are no relative maxima.

Explanation:

Step1: Find the first - order partial derivatives

The function is \(f(x,y)=2x^{2}+3xy + 4y^{2}-6x + 7y\).
The partial derivative with respect to \(x\) is \(f_{x}=\frac{\partial f}{\partial x}=4x + 3y-6\).
The partial derivative with respect to \(y\) is \(f_{y}=\frac{\partial f}{\partial y}=3x + 8y+7\).

Step2: Set the first - order partial derivatives equal to zero

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:

$$
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$$

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 the second - order partial derivatives

\(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}} = 4\), \(f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=3\), \(f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=8\).
The discriminant \(D=f_{xx}f_{yy}-(f_{xy})^{2}\).
Substitute the values: \(D=(4\times8)-3^{2}=32 - 9=23>0\) and \(f_{xx}=4>0\).

Answer:

A. The point(s) is(are) \((3,-2)\)