find all the local maxima, local minima, and saddle points of the given function.\n\nf(x,y)=2x² + 3xy + 4y²…

Explanation:

Step1: Find first - order partial derivatives

The first - order partial derivatives of (t(x,y)=2x^{2}+3xy + 4y^{2}-6x + 7y) are: (t_{x}=\frac{\partial t}{\partial x}=4x + 3y-6) (t_{y}=\frac{\partial t}{\partial y}=3x + 8y+7)

Step2: Solve the system of equations (t_{x}=0) and (t_{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: (3\times\frac{6 - 3y}{4}+8y+7 = 0) (\frac{18-9y}{4}+8y+7 = 0) Multiply through by 4: (18-9y + 32y+28 = 0) (23y=-46), so (y = - 2) Substitute (y = - 2) into (x=\frac{6 - 3y}{4}), we get (x=\frac{6-3\times(-2)}{4}=\frac{6 + 6}{4}=3)

Step3: Find second - order partial derivatives

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

Step4: Use the second - derivative test

The discriminant (D=t_{xx}t_{yy}-(t_{xy})^{2}) (D=(4\times8)-3^{2}=32 - 9=23>0) and (t_{xx}=4>0)

Answer:

B. There are no local maxima.