Lagrange multiplier theorem has significantly reduced the difficulty of finding solutions to constraint problems. It is used to find maxima/minima of a function within a given constraint. %%visits: 5 # diary long, and I don’t feel like this is where I want it to be in terms of “done” # intuition :~ finding an extremum of a function within another “constraint” function. The highest point of a mountain inside the clouds. at this point the vectors of mountains = K vectors of clouds
When finding the minimum or maximum of f at a constraint g. # rigour := If the function f(x) has an extremum (subject to g(x) = c, g(x) is the constraint) at the point x0, then the vectors ∇f(x0) and ∇g(x0) and parallel. Thus if ∇g(x0) is non-zero, there exists a unique constant λ (known as a Lagrange multiplier) such that ∇f(x0) = λ∇g(x0) # exam clinic 1. Use the method of Lagrange multipliers to find the distance of the origin from the plane. 2x − 2y + z = 5 - The Lagrange formula above has f(x) and g(x). First we need to know the main function and the constraint. The main function is not given to us but we can find it. The modulus function is $\sqrt{x^2+y^2+z^2}$, so that is f(x). Now to work out ∇f and ∇g. $\nabla f = \lambda \left( x\left( x^2+y^2+z^2 \right)^{-\frac{1}{2}}\hat{i} + y\left( x^2+y^2+z^2 \right)^{-\frac{1}{2}}\hat{j} + z\left( x^2+y^2+z^2 \right)^{-\frac{1}{2}}\hat{k} \right)$ λ∇g = λ(2î − 2ĵ + k̂) $\frac{x}{f(x)} = 2\lambda$ $\frac{y}{f(x)} = -2\lambda$ $\frac{z}{f(x)} = \lambda$ Putting all the ’λ’s together gives the answer. # resources https://www.youtube.com/watch?v=8mjcnxGMwFo tags :math:calculus_2: