Posted by Anonymous on Monday, April 4, 2011 at 9:50am.
"Using Lagrange multipliers, find the maximum value of f(x,y) = x + 3y + 5z subject to the constraint x^2 + y^2 + z^2 = 1."
Any help would be appreciated!

Calculus  MathMate, Monday, April 4, 2011 at 4:55pm
This can be solved by geometry.
The constraint is the surface of a sphere of radius 1.
The given plane has a normal unit vector of (i+3j+5k)/sqrt(1²+3²+5²)
=(i+3j+5k)/sqrt(35).
So the maximum and minimum value of x+3y+5z is at
(x,y,z)=(1,9,25)/sqrt(35).
Which when substituted into the equation of the plane gives P(x,y,z)=1.
Using Lagrange multipliers:
Objective function:
P(x,y,z)=x+3y+5z+L(x²+y²+z²1)
where L=lagrange multiplier (lambda)
Partially differentiate with respect to x, y and z gives the first order conditions:
∂P/∂x = 1+2xL = 0
∂P/∂y = 3+2yL = 0
∂P/∂z = 5+2zL = 0
Solve for x,y and z in terms of L and substitute in the constraint equation of x²+y²+z²=1
(1/2L)²+(3/2L)²+(5/2L)² = 1
Solve for L to get
L=±sqrt(35)/2
Substitute to get maximum
x= 1/2L = 1/sqrt(35)
y= 3/2L = 3/sqrt(35)
z= 5/2L = 5/sqrt(35)
or
P(x,y,z)=(1+9+25)/sqrt(35)=1