Example of Constrained Optimization

Find the maximum of the function f(x,y) = x² - y subject to the constraint that $g(x,y) = x^2 + y^2 \le c^2$.

1.

Inside the circular constraint region, $\mbox{grad} f = 2x \hat{i} - \hat{j} = \vec{0}$ gives an entire line of critical points: (0, y) where $-c \le y \le c$ since we must stay inside the region.

2.

On the boundary, we must solve the equations

$$\mbox{grad}f = \lambda \mbox{grad}g \qquad \qquad g(x,y) = c^2.$$ (11)

These can be reformulated as

$$2x = 2 \lambda x \qquad \qquad -1 = 2 \lambda y \qquad \qquad x^2 + y^2 = c^2.$$ (12)

The leftmost equation says that $2x(1 - \lambda) = 0$ which admits two solutions, x = 0 and $\lambda = 1$.If we take x = 0 the third equation gives $y = \pm c$ and the then we have $\lambda = \mp 1/2c$.

If we take $\lambda = 1$ the second equation gives y = -1/2 and the third then gives us that $x = \pm \sqrt{c^2 - 1/4}$.

Thus we have located four critical points on the boundary

$$(0, \pm c, \lambda = \mp 1/2c) \qquad \qquad (\pm\sqrt{c^2 - 1/4},-1/2).$$ (13)

3.

We now compute the value of f at each critical point.

(a)

f(0,y) = -y. Since this critical point must lie inside the region, we know that -y lies between -c and +c.

(b)

$f(0,\pm c) = \mp c$

(c)

$f(\pm \sqrt{c^2 - 1/4}, -1/2) = c^2 + 1/4$

Thus, the maxima are clearly the points $(\pm \sqrt{c^2 - 1/4},-1/2)$.