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

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: [ \begin{align*} 3\times\frac{6 - 3y}{4}+8y+7&=0\ \frac{18-9y}{4}+8y + 7&=0\ 18-9y+32y+28&=0\ 23y&=-46\ y&=- 2 \end{align*} ] 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))