Find all the local maxima, local minima, and saddle points of the function.
-$$f ( x , y ) = x ^ { 2 } + 14 x + y ^ { 2 } - 4 y + 3$$
A) f(7, -2) = 162, local maximum
B) f(-7, 2) = -50, local minimum
C) f(-7, -2) = -34, saddle point
D) f(7, 2) = 146, saddle point

Question

Quizplus · Accepted Answer

To find the local maxima, local minima, and saddle points of the function $f(x, y) = x^2 + 14x + y^2 - 4y + 3$, we first find its critical points by taking the partial derivatives with respect to $x$ and $y$, setting them equal to zero, and solving for $x$ and $y$.The partial derivative with respect to $x$ is $f_x = 2x + 14$, and setting it to zero gives $x = -7$.The partial derivative with respect to $y$ is $f_y = 2y - 4$, and setting it to zero gives $y = 2$.Thus, the only critical point is $(-7, 2)$.To classify this critical point, we examine the second partial derivatives: $f_{xx} = 2$, $f_{yy} = 2$, and $f_{xy} = f_{yx} = 0$. The determinant of the Hessian matrix at the critical point is $D = f_{xx}f_{yy} - (f_{xy})^2 = 2*2 - 0 = 4$, which is positive. Since $f_{xx} > 0$, the critical point is a local minimum.Therefore, the correct answer is that $(-7, 2)$ is a local minimum, and the value of the function at this point is $f(-7, 2) = (-7)^2 + 14*(-7) + (2)^2 - 4*2 + 3 = 49 - 98 + 4 - 8 + 3 = -50$.

Find All the Local Maxima, Local Minima, and Saddle Points f(x,y)=x2+14x+y2−4y+3f ( x , y ) = x ^ { 2 } + 14 x + y ^ { 2 } - 4 y + 3f(x,y)=x2+14x+y2−4y+3

Find All the Local Maxima, Local Minima, and Saddle Points $f ( x , y ) = x ^ { 2 } + 14 x + y ^ { 2 } - 4 y + 3$