dynamicsolo said:I withdraw my previous remark: the Lagrange equation has an x2 term which equals a y2 term , so the critical points are symmetrical about the origin, but there are only two (not four as I earlier proposed). Since the function has an elliptic paraboloid for a surface, z is only ≥ 0 , so there won't be maxima. The constraint has a surface which is a hyperbolic cylinder, so the intersection curves turn out to be "upward-opening" parabolas. Calculate the "D-index" (the thingie with the second partial derivatives) to satisfy yourself that the critical points are minima.
dynamicsolo said:I withdraw my previous remark: the Lagrange equation has an x2 term which equals a y2 term , so the critical points are symmetrical about the origin, but there are only two (not four as I earlier proposed). Since the function has an elliptic paraboloid for a surface, z is only ≥ 0 , so there won't be maxima. The constraint has a surface which is a hyperbolic cylinder, so the intersection curves turn out to be "upward-opening" parabolas. Calculate the "D-index" (the thingie with the second partial derivatives) to satisfy yourself that the critical points are minima.
Dick said:Oooh. Now I deleted my previous post because I felt kind of doubtful. Notice you can actually solve for y=5/x. Substitute that in for y and find the extrema as a single variable problem to check your Lagrange solution.
Dick said:Oooh. Now I deleted my previous post because I felt kind of doubtful. Notice you can actually solve for y=5/x. Substitute that in for y and find the extrema as a single variable problem to check your Lagrange solution.
dynamicsolo said:Sorry to have dropped a small spanner into the works. When the equations for [itex]\lambda[/itex] led me to 4x2 = 6y2 , I immediately thought "four-fold symmetry". But the constraint equation imposes a diagonal symmetry, so the number of critical points is forced to be only two, rather than four...
Lagrange multipliers are a mathematical technique used to find the minimum or maximum value of a function subject to a set of constraints. This allows for optimization problems to be solved in a more efficient manner.
To find the minimum and maximum values of a function using Lagrange multipliers, you must first set up the Lagrangian function by adding the function to be optimized to the product of the constraints and their corresponding Lagrange multipliers. Then, take the partial derivatives of the Lagrangian function with respect to each variable and set them equal to 0. Solve the resulting system of equations to find the values of the variables that correspond to the minimum or maximum value.
The constraints in Lagrange multipliers are the conditions that must be satisfied in order to find the minimum or maximum value of the function. These constraints can be in the form of equations or inequalities.
Yes, Lagrange multipliers can be used for functions with multiple variables. In this case, the Lagrangian function will have multiple constraints and corresponding Lagrange multipliers, and the partial derivatives will need to be taken with respect to each variable.
Lagrange multipliers have many real-world applications, including in economics, physics, and engineering. They can be used to optimize production processes, minimize costs, and find the optimal shape or dimensions of objects. They are also useful in solving optimization problems in finance and in finding equilibrium points in game theory.