calculus 3 help, not sure if I posted in right place

**Noobie** · April 18th 2009, 11:42 AM

Hi guys,

Like I said Im not sure If I posted this in the right spot, but if someone can take a look at these 3 problems and help me out I would appreciate it.

1.) Find ther extreme values of the function subject to the given constraint. F(x,y,z)= x^3 + y^3+z^3, x^2+y^2+z^2=4

2.)Write parametric equations for the tangent line to the curve of intersection of the surfaces. x+y^2+7z=9 amd x=1 at the point (1,1,1)

3.) A normal line tot he paraboloid z=6x^2 +2y^2 also passes thorugh the point (26,25,73). Find the point on the paraboloid that the normal line passes through.

any help would be greatly appreciated, this is a test review for our final exam and Ive got all the other material figured out, and have spent 3 days and a crap load of time and paper trying to solve these but am havin no luck.

**Danny** · April 18th 2009, 12:22 PM

Originally Posted by Noobie

Hi guys,

Like I said Im not sure If I posted this in the right spot, but if someone can take a look at these 3 problems and help me out I would appreciate it.

1.) FIND THE EXTREME VALUES OF THE FUNCTION SUBJECT TO THE GIVEN CONSTRAINT. F(x,y,z)= x^3 + y^3+z^3, x^2+y^2+z^2=4

2.)Write parametric equations for the tangent line to the curve of intersection of the surfaces. x+y^2+7z=9 amd x=1 at the point (1,1,1)

3.) A normal line tot he paraboloid z=6x^2 +2y^2 also passes thorugh the point (26,25,73). Find the point on the paraboloid that the normal line passes through.

any help would be greatly appreciated, this is a test review for our final exam and Ive got all the other material figured out, and have spent 3 days and a crap load of time and paper trying to solve these but am havin no luck.

Here's a bit on the first. Using Lagrange multipliers

$G = x^3+y^3+z^3 + \lambda \left( x^2+y^2+z^2-4\right)$

$G_x = 3x^2 + 2 x \lambda$ , $G_y = 3y^2 + 2 y \lambda$ ,
$G_z = 3z^2 + 2 z \lambda$ , $G_{\lambda} = x^2+y^2+z^2-4$

which we set all to zero. From the first three we obtain

$x = 0,\;\;\text{or}\;\;x = - \frac{2 \lambda}{3}$
$y = 0,\;\;\text{or}\;\;y = - \frac{2 \lambda}{3}$
$z = 0,\;\;\text{or}\;\;z = - \frac{2 \lambda}{3}$

which gives rise to the following three cases

(i) one of $x,\; y, \; \text{or}\; z$ is zero, the other two $- \frac{2 \lambda}{3}$ or equal to each other
(ii) two of $x,\; y, \; \text{or}\; z$ is zero the other $- \frac{2 \lambda}{3}$
(iii) $x = y = z = - \frac{2 \lambda}{3}$ or all equal to each other

(i) Say z = 0, so $x = y$ and from the constraint $x = y =\pm \sqrt{2}$ so $F = \pm \frac{8}{\sqrt{2}}$
(ii) Say y = z = 0, so from the constraint $x = \pm 2$ so $F = \pm 8$
(iii) $x = y = z = \pm \frac{2}{\sqrt{3}}$ giving $F = \pm \frac{8}{\sqrt{3}}$

You could also use

$x = 2 \cos \theta \sin \phi, y = 2 \sin \theta \sin \phi, z = 2 \cos \phi,$ as this satisfies the constaint but it might be a bit messy.

**Noobie** · April 18th 2009, 12:38 PM

that actually does kinda help with the first one, thanks bro

**Noobie** · April 19th 2009, 02:09 PM

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