The set of points on $z = x ^ { 2 } - 3 x y + y ^ { 2 }$ at which the tangent plane is parallel to the xy-plane is

A) $\{ ( 0,0,3 ) \}$
B) $\{ ( 0,0,2 ) \}$
C) $\{ ( 0,0,1 ) \}$
D) $\{ ( 0,0,0 ) \}$
E) $\{ ( 0,0,5 ) \}$

Question

The set of points on $z = x ^ { 2 } - 3 x y + y ^ { 2 }$ at which the tangent plane is parallel to the xy-plane is

A) $\{ ( 0,0,3 ) \}$
B) $\{ ( 0,0,2 ) \}$
C) $\{ ( 0,0,1 ) \}$
D) $\{ ( 0,0,0 ) \}$
E) $\{ ( 0,0,5 ) \}$

Quizplus · Accepted Answer

The tangent plane to the surface $z = f(x,y)$ at a point $(x_0, y_0, z_0)$ is given by $z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$, where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively. For the plane to be parallel to the xy-plane, both partial derivatives must be zero at $(x_0, y_0)$. Given $f(x, y) = x^2 - 3xy + y^2$, we find $f_x = 2x - 3y$ and $f_y = -3x + 2y$. Setting these equal to zero, we solve $2x - 3y = 0$ and $-3x + 2y = 0$ simultaneously, yielding $x = 0$ and $y = 0$. Substituting $x = 0$ and $y = 0$ into the original equation gives $z = 0$, so the point is $(0,0,0)$.

The Set of Points On z=x2−3xy+y2z = x ^ { 2 } - 3 x y + y ^ { 2 }z=x2−3xy+y2

The Set of Points On $z = x ^ { 2 } - 3 x y + y ^ { 2 }$