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