Question

Among all points on the graph of z=12-x^2-y^2 that lies above the plane x+3y+6z=0, find the point farthest from the plane.

Answer 1

+10xp!

Answer 2

A point that lies on the graph of\[z = 12-x^2-y^2\]is a point of the form\[\left(x, y, 12-x^2-y^2\right).\]If it also lies above the plane\[x + 3y + 6z,\]it means that \[12-x^2-y^2 > -\frac{1}{6}(x + 3y) \Rightarrow12-x^2-y^2+\frac{1}{6}(x+3y)>0.\]
The distance between this point and the plane is\[\overline{D}(x,y) = \frac{x + 3y + 6(12 - x^2 - y^2)}{\sqrt{46}}.\]We can forget the root in the denominator and define\[D(x, y) \doteq x + 3y -x^2 - y^2 + 72.\]If we set\[f(x, y) \doteq 12 - x^2 - y^2 + \frac{x}{6} + \frac{y}{2} = a > 0\]we can use Lagrange multipliers to find a point\[(x_a,y_a)\]such that\[D(x_a,y_a) = \max_{f(x,y) = a} D(x,y):\]\[\Lambda(x,y,\lambda) = D(x,y) + \lambda(f(x,y)-a)\]\[\frac{\partial}{\partial x}\Lambda(x,y,\lambda) = \frac{6+\lambda}{6}-2x(1+\lambda)=0\]\[\frac{\partial}{\partial y}\Lambda(x,y,\lambda) = \frac{1}{2}(6+\lambda-4y(1+\lambda))=0\]\[\frac{\partial}{\partial \lambda}\Lambda(x,y,\lambda) = \frac{1}{6}\left(72-6a+x-6x^2+3y-6y^2\right)=0\]Solving these equations we get\[x_a = \frac{1}{60}\left(5\pm\sqrt{5(869-72a)}\right)\]\[y_a = \frac{1}{20}\left(5\pm\sqrt{5(869-72a)}\right)\]and\[D(x_a,y_a) = 6a\]so the distance is at a maximum when a is at a maximum. Since a is the distance along the Z axis between the paraboloid and the plane (which is also f(x, y)), a is at a maximum when f(x, y) is at a maximum and the problem becomes finding the maximum of f(x, y).
\[\frac{\partial}{\partial x}f(x, y) = \frac{1}{6} - 2x = 0 \Rightarrow x = \frac{1}{12}\]\[\frac{\partial}{\partial y}f(x, y) = \frac{1}{2} - 2y = 0 \Rightarrow y = \frac{1}{4}\]\[z = 12 - x^2 - y^2 = \frac{869}{72}\]so the point farthest from the plane is\[\left(\frac{1}{12}, \frac{1}{4}, \frac{869}{72}\right).\]

Answer 3

Thank you so much

Answer 4

What math level is this?

Answer 5

Multivariable calculus. Typically college calculus 3