Amy Blackburn
MAT 211 HONORS
March 11, 2014
Professor Brewer
The History of Lagrange Multipliers
When finding the relative extrema of an objective function, there can be constraints on the independent variables. Lagrange multipliers allow you to find the local maxima and minima of a function with constraints. It is used in place of substitution when it is too difficult to solve a constraint equation for one variable. To find the candidates for relative extrema of the function f(x,y,…) subject to the constraint g(x,y,…) you use the method of Lagrange multipliers.
Joseph- Louis Lagrange, the man who discovered this method was born in northern Italy in 1736. He was one of the founders of variable calculus and was a professor at age 19. By age 25 he was known as the greatest mathematician alive. He was the director of the Berlin Academy and then was invited to the Paris Academy of Science by Louis XVI. Joseph- Louis Lagrange died in 1813 and was buried in the Pantheon.
To find the relative maximum and minimums of f(x,y,…) subject to the constraint g(x,y…)=0 you must use the Lagrangian function. The Lagrangian function is:
L(x,y,…) = f(x,y,…) –λ g(x,y,…)
λis a unknown called a Lagrange multiplier.
To solve, start by plugging in all components into the formula given above. f(x,y,…) and g(x,y,…) are generally given in the question. Then you need to solve for Lx =0 and Ly=0 by taking the partial derivative of x or y accordingly and set them equal to zero. You will also need to set the constraint equal to zero. These three equations will form our system of equations.
From there use substitution to solve one of the equations for λ. Once you have λ you can plug it into the other equations and solve for x and y. This will give you ordered pairs which are your possible maximum or minimum. To find the maximum, plug the points back into f(x,y,…) and choose the ordered pair that produces the highest value.
Consider the following problem:
Find the maximum value of f(x,y,z) = 1-x2-x-y2+y-z2+z subject to 3x=y
L=1-x2-x-y2+y-z2+z - λ(y-3x)
Lx=0: 2x-1+λ3=0
Ly=0: 2y-1 -λ=0
Lz=0: 2z+1=0
g=0: y-3x=0 λ=-2y+1
2x-1+ (-2y+1)3=0 2x-1-6y+3 2x-6y+2=0 x=3y-1
g=0: y-3(3y-1)=0 y=3/10
3/10-3x=0 x=1/10
Lz=0: 2z+1=0 z=1/2
f(x,y,z) = 1-x2-x-y2+y-z2+z (1-(1/10)2-(1/10)-(3/10)2+(3/10)-(0.5)2+(0.5) 1.35
To better explain, I will provide a real life example:
Two different manufacturers produce cars at factories A and B. Suppose “x” cars are made at A and “y” cars are made at B. Suppose the cost is given by f(x,y)= 4x2+8y2
If you must produce 60 cars, what is the number you should produce at each factory?
Constraint: x+y=60
L: 4x2+8y2-λ(x+y-60)
Lx= 8x- λ=0 x=λ/8
Ly=16y-λ=0 y=λ/16
Lλ= -x-y+60=0 Plug in above values to this equation
Lλ= -(λ/8) – (λ/16) +60 = 0
-(3λ)/16 = -60 λ/16= 20 λ= 320 Plug in λ to Lx Ly equations.
x= 40
y= 20
In conclusion, using Lagrange multipliers is easy way to find all of the critical points on a surface or curve. Using these critical points we can evaluate the equation. The largest number found is the maxima and the smallest is the minima. Lagrange theorem is a useful way to find the maxima and minima of a surface or curve while excluding any unnecessary areas.
Works Cited
Luigi, Giuseppe. Joseph-Louis.Digital image.Wikipedia.N.p., n.d. Web. 21 April 2014.