Calc III Tutorial: Lagrange Multipliers

3 months ago\({\bf{Purpose:}}\) to find local extrema when a function f is constrained by another function g ex: (without using multipliers) on the plane 2x + y - z - 5 = 0 what point is closest to the origin? here, the constraint function is the plane, since that limits what our points can be the function we want to maximize or minimize is \[\sqrt{x^{2}+y^{2}+z^{2}}\] or the distance equation, in other words. we can take the original function 2x + y - z - 5 = 0 and write it as a function of z z = 2x + y - 5 which allows us to re-write the distance function as \(\sqrt{x^{2}+y^{2}+(2x + y - 5 )^{2}}\) now take the partial derivatives wrt x and y you should know how to take partial derivatives by now so I'll skip some of the algebraic steps fx = 10x + 4y - 20 fy = 4x + 4y - 10 = 0 from there it should be straightforward to solve this system for both x, and y, and plug these back in the original eqn to find z.

3 months ago\({\bf{Method}}\): of course, the above method is a bit tedious and doesn't work for all functions so we will introduce another method of doing this ∇f = λ∇g for a function f constrained by a surface g = 0 Orthogonal Gradient Theorem for a smooth curve C = r(t) = x(t)i + y(t)j + z(t)k containing point P0 where f has a local min or max with respect to the values of f on C, then ∇f is orthogonal to C at P0, or in other words ∇f dot r' = 0

3 months ago\({\bf{Example}}\) f(x,y) = xy on the ellipse x^2/8 + y^2/2 = 1 ∇f = yi + xj ∇g = (λ/4)xi + λyj setting the i components equal to each other, and the j components equal to each other y = (λ/4)x x = λy plugging the solution to x into the y equation gives us y = (λ/4)*λy, or just (λ^2/4)y this equation is only satisfied when 1) y = 0 or 2) λ = +/-2 if y = 0 then via x = λy, x must also be 0, however, these conditions do not satisfy the original ellipse equation. if λ = +/-2 then x = +/-2y you can plug this back into the original ellipse equation to solve for y, will skip some algebraic steps but you should end up with y = +/- 1 so going back to the original f(x,y) = xy function the extremes are 2 and -2

3 months agoSource material is section 14.8 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al. there's one particular QC troll on the verge of being banned, or just leaving, so I might be able to leave tutorials open for once without getting harassed or spammed.

3 months ago