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Posted by 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 - , 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+z²-1)
    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

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