Constrained extrema

You solve the system by setting up the equations $\nabla f = \lambda \nabla g$ alongside the constraint $g(x,y)=0$. For these specific functions, this yields $2x = \lambda(2x)$, $2y = \lambda(y)$, and $x^2 + y^2/2 - 1 = 0$.

Conditions: The objective function is $f(x,y) = x^2 + y^2$.; The constraint function is $g(x,y) = x^2 + y^2/2 - 1 = 0$.; The Lagrange multiplier condition $\nabla f = \lambda \nabla g$ is applied.

The blue arrows represent the gradient of the objective function, $\nabla f$, pointing outward from the center of the level curves. The yellow arrows represent the gradient of the constraint function, $\nabla g$, which is normal to the constraint curve.

Conditions: The animation visualizes the gradients of $f(x,y) = x^2 + y^2$ and $g(x,y) = x^2 + y^2/2 - 1 = 0$.; Blue arrows correspond to $\nabla f$.; Yellow arrows correspond to $\nabla g$.

The video uses the objective function $f(x,y) = x^2 + y^2$, which is represented as a 3D paraboloid with concentric circular level curves. The constraint function is $g(x,y) = x^2 + y^2/2 - 1 = 0$, which is represented as an ellipse on the xy-plane.

Conditions: The demonstration focuses on constrained optimization in two variables.; The objective function is quadratic and radially symmetric.; The constraint is an ellipse defined by a quadratic equation.

At the extremum points, the level curves of the objective function are tangent to the constraint curve. Because gradient vectors are always perpendicular to their respective level curves, the gradients of the objective and constraint functions must be parallel at these points of tangency.

Conditions: The functions $f$ and $g$ are differentiable.; The extremum occurs at a point where the constraint curve is smooth (non-zero gradient for $g$).; The level curves of $f$ are tangent to the constraint curve $g=0$.