How to find point on surface closest to the origin of 9z = x2 + 3y − 45?
How to find the point on surface closest to the origin of 9z = x2 + 3y − 45?
I know I need to Minimize the square of the distance from (x, y, z) to the origin but I can't seem to get the right coordinates. I know x=0, but that's about it.
Any help would be appreciated.
Update:Oops. I meant
9z = x^2 3y − 45
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It suffices to minimize the square of the distance (call in D) to the origin:
D = (x - 0)^2 + (y - 0)^2 + (z - 0)^2
...= (-3y + 9z + 45) + y^2 + z^2, since 9z = x^2 + 3y - 45.
Taking partial derivatives:
D_y = -3 + 2y and D_z = 2z + 9.
Setting these equal to 0 yields the critical point (y, z) = (3/2, -9/2).
Since D_yy = 2 > 0, and (D_yy)(D_zz) - (D_yz)^2 = (2)(2) - 0^2 = 4 > 0,
we have a local minimum at (y, z) = (3/2, -9/2).
[In fact, it is a global minimum, since we can complete the square on D:
D = (y^2 - 3y) + (z^2 + 9z) + 45
...= (y^2 - 3y + 9/4) + (z^2 + 9z + 81/4) + 45 - 9/4 - 81/4
...= (y - 3/2)^2 + (z + 9/2)^2 + 45/2.
The sum of squares is minimal when both squares equal 0 <==> (y, z) = (3/2, -9/2).]
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Since 9z = x^2 + 3y - 45, we have the point (x, y, z) = (0 , 3/2, -9/2).
I hope this helps!