Solve the problem.
-Find the point on the paraboloid $z = 2 - x ^ { 2 } - y ^ { 2 }$ that is closest to the point $( 1,1,2 )$.
A) $\left( - \frac { 1 } { 2 } , - \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)$
B) $\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)$
C) $\left( - \frac { 1 } { 2 } , \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)$
D) $\left( \frac { 1 } { 2 } , - \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)$

Question

Quizplus · Accepted Answer

To find the point on the paraboloid $z = 2 - x^2 - y^2$ that is closest to the point $(1,1,2)$, we can use the distance formula in three dimensions. However, because we are looking for a minimum distance, it's more efficient to minimize the square of the distance to avoid dealing with square roots. The square of the distance $D^2$ between a point $(x, y, z)$ on the paraboloid and the point $(1,1,2)$ is given by:$$D^2 = (x-1)^2 + (y-1)^2 + (z-2)^2$$Substituting $z = 2 - x^2 - y^2$ into the equation gives:$$D^2 = (x-1)^2 + (y-1)^2 + (2 - x^2 - y^2 - 2)^2$$$$= (x-1)^2 + (y-1)^2 + (-x^2 - y^2)^2$$To minimize $D^2$, we take the partial derivatives with respect to $x$ and $y$, set them equal to zero, and solve for $x$ and $y$.$$\frac{\partial D^2}{\partial x} = 2(x-1) - 2x = 0$$$$\frac{\partial D^2}{\partial y} = 2(y-1) - 2y = 0$$Solving these equations gives $x = \frac{1}{2}$ and $y = \frac{1}{2}$. Substituting these values back into the equation for $z$ gives $z = 2 - (\frac{1}{2})^2 - (\frac{1}{2})^2 = \frac{3}{2}$.Therefore, the point on the paraboloid closest to $(1,1,2)$ is $\left(\frac{1}{2}, \frac{1}{2}, \frac{3}{2}\right)$.

Solve the Problem z=2−x2−y2z = 2 - x ^ { 2 } - y ^ { 2 }z=2−x2−y2

Solve the Problem $z = 2 - x ^ { 2 } - y ^ { 2 }$