Optimization Using Lagrange Multipliers

Optimization is selecting the best in the given situation.  Calculus can be used in optimising many problems like selection of least material to manufacture something or selection of the method that costs least to manufacture.  Optimization involving single variable is simple and done by finding critical points, boundary points and discontinuous or nondifferentiable points.  In the same way problems involving two or more variables can also be optimized.  It is done by putting some given condition in the function which we are going to optimise and find points that were said in case of single variable optimization.  But in case of multivariable functions we equal all first order partial derivatives to zero.

We are doing here an optimization problem but not by conventional method we used to do in single variable optimization.  We are using another method called Lagrange multipliers.  This process of optimization may make optimization process easier in some of the cases.  So it is useful method of optimization to know.



We are manufacturing a cylinder can with both the sides closed.  We want to manufacture it for fixed volume using the least amount of material.  So we are optimizing the surface area of the can for the fixed volume.

Let f(r,h) denote the function to be optimised = surface area function which depends on two variables i.e radius of can(r) and height of the can(h).  And g(r,h) be the volume function which also depends on the same variables.  As the volume is fixed, g(r,h) = V is a level set of the volume function.  We are required here to optimise the surface area function under the level set g(r,h)=V.

The surface area function is given as:

And the level set is given as:

Now let’s find gradient vector function of both volume as well as surface area functions.

Gradient vector function of surface area function,

Similarly gradient vector function of volume function,

Here gradient vector function of volume function can only be zero at r=0 and h=0 which has no volume so not taken this value into consideration.
So gradient vector function of volume function can never be zero according to our condition.

Therefore,

Taking equation (2), we cancel 'r' from both sides as r can't be zero.  Then the value of ‘λ’ is given as:

Putting this value of ‘λ’ in (1),

So, h = 2r is the condition of optimization.  The can can be made by least amount of material if we make it of shape such that diameter of base equals height of the can.

Putting this condition in the level set g(r,h) = V we can find corresponding dimensions of the can. In this way Lagrange multipliers can be used to optimize the multivariable functions.  We saw above how easily we worked this optimization problem with this method.