# Math Help - Help!!! find range, domain, max point on surface!!

1. ## Help!!! find range, domain, max point on surface!!

Hello, I was wondering if you guys can help me understanding this problem.

f(x,y)=(11-x^2-y^2+4*x+2*y)^(1/2)

Find: Identify the surface, domain, range, plot the surface,and find the maximum point on the surface!

I know how to find the max point, if it's in 2D , just take the derivative and set it equal to zero. But how can I do it in 3D?

[ for plotting, I am not sure, If I plot it in x, y I get the upper half of an ellipse, that mean in 3D it will be quadrant of an ellipse shape! ]

2. Originally Posted by beho86
Hello, I was wondering if you guys can help me understanding this problem.

f(x,y)=(11-x^2-y^2+4*x+2*y)^(1/2)

Find: Identify the surface, domain, range, plot the surface,and find the maximum point on the surface!

I know how to find the max point, if it's in 2D , just take the derivative and set it equal to zero. But how can I do it in 3D?

[ for plotting, I am not sure, If I plot it in x, y I get the upper half of an ellipse, that mean in 3D it will be quadrant of an ellipse shape! ]
You certainly can complete the square, set it equal to z and square to get something like $(x- a)^2+ (y- b)^2+ z^2= r^2$ so, yes, this is one quadrant of an ellipsoid. You can also get the domain and range from that. And, having recognized that it is an ellipsoid, you should be able to see that the maximum value of f(x,y), the z value, is directly above the center point of the ellipsoid- that would at x= a, y= b as given above- or simply recognize that $f(x,y)= \sqrt{r^2- (x- a)^2+ (y-b)^2}$ will be maximum when x= a and y= b because otherwise you are subtracting something from $r^2$.

3. Thanks HallsofIvy,

So, That's correct?

the domain is all real x, y, z such that (11-x^2-y^2+4*x+2*y) >= 0
Range: f(x,y) >= 0

11-x^2-y^2+4*x+2*y
= - (X - 2)2 + 4 - (Y - 1)2 + 1 + 11 =..... 16
So, Max point is (2, 1, 4 )