Question

Use Lagrange multipliers to find the minimum value of the function, f(x,y)= x2 + y2 subject to the constraint x + y =1

Answer 1

The Lagrangian is

`L(x,y,\lambda) = x^2 + y^2 - \lambda (x + y - 1)`

Find its critical points.

`(\partial L)/(\partial x) = 2x - \lambda = 0 \implies x = \frac\lambda2`

`(\partial L)/(\partial y) = 2y - \lambda = 0 \implies y = \frac\lambda2`

`(\partial L)/(\partial\lambda) = -(x+y-1) = 0 \implies x + y = 1`

From the first two conditions,
`x=y`, so

`x + y = 2x = 1 \implies x = y = \frac12`

At the point (1/2, 1/2), compute the Hessian determinant of
`f(x,y)`.

`H(x,y) = \begin{bmatrix} (\partial^2f)/(\partial x^2) & (\partial^2f)/(\partial x \partial y) \\\\ (\partial^2f)/(\partial y\partial x) & (\partial^2f)/(\partial y^2) \end{bmatrix} = \begin{bmatrix}2&0\\0&2\end{bmatrix}`

`H(x,y)` is positive definite for all
`x,y`, which indeed indicates a local minimum at (1/2, 1/2), so that

`\min\left\{x^2+y^2 \mid x+y=1\right\} = f\left(\frac12,\frac12\right) = \frac14+\frac14 = \boxed{\frac12}`