Examine the function $f ( x , y ) = 4 x ^ { 2 } - 2 x y + 4 y ^ { 2 } + 120 x + 60 y$ for relative extrema.
A)relative $\text { minimum }$ at $( - 18 , - 12 )$
B)relative $\text { maximum }$ at $( - 18 , - 12 )$
C)relative $\text { minimum }$ at $( - 18,12 )$
D)relative $\text { maximum }$ at $( - 18,12 )$
E)no relative extrema

Question

Quizplus · Accepted Answer

To find relative extrema, we first find the partial derivatives of $f(x, y)$ with respect to $x$ and $y$, set them equal to zero, and solve for $x$ and $y$.$$f_x = \frac{\partial f}{\partial x} = 8x - 2y + 120$$$$f_y = \frac{\partial f}{\partial y} = -2x + 8y + 60$$Setting $f_x = 0$ and $f_y = 0$ gives us a system of equations:$$8x - 2y + 120 = 0$$$$-2x + 8y + 60 = 0$$Solving this system yields $x = -18$ and $y = -12$.To determine whether this point is a maximum, minimum, or saddle point, we examine the second partial derivatives and the determinant of the Hessian matrix $D$.$$f_{xx} = 8, f_{yy} = 8, f_{xy} = f_{yx} = -2$$$$D = f_{xx}f_{yy} - (f_{xy})^2 = (8)(8) - (-2)^2 = 64 - 4 = 60$$Since $D > 0$ and $f_{xx} > 0$, the function has a relative minimum at $(-18, -12)$.

Examine the Function f(x,y)=4x2−2xy+4y2+120x+60yf ( x , y ) = 4 x ^ { 2 } - 2 x y + 4 y ^ { 2 } + 120 x + 60 yf(x,y)=4x2−2xy+4y2+120x+60y

Examine the Function $f ( x , y ) = 4 x ^ { 2 } - 2 x y + 4 y ^ { 2 } + 120 x + 60 y$