Find the extreme values of the function: $z=f(x,y)=x^2+(y-18)^2+90$
subjected to following constraint
$x^2+y^2\leq196$
How to solve this? I used Lagrangian function but how to set up constraint qualification? Do we have to assume that $x,y\geq0$
Which would be the best method to solve this?
On
We have: $\left(\dfrac{x}{14}\right)^2 + \left(\dfrac{y}{14}\right)^2 \leq 1$. Thus we can take $\dfrac{x}{14} = r\cos \theta, \dfrac{y}{14} = r\sin \theta$ with $0 \leq r \leq 1$. Thus $z = f(r,\theta) = 196r^2\cos^2 \theta + \left(14r\sin \theta - 18\right)^2 + 90= 196r^2 - 504r\sin \theta + 414$. Because $-1 \leq \sin \theta \leq 1$ we have: $z \leq 196r^2 +504r + 414$. You can continue to find the max of this function on $[0,1]$,and similarly can do the same for $z \geq 196r^2 - 504r + 414$, and you can find the minimum of this function on $[0,1]$. Together with the max you have both extreme values of $z = f(x,y)$.
Basically $z$ is equal to the square of the distance of (0,18) from any point inside the circle plus 90.
Hence while minimising this distance we would take the point (0,14) and while maximising it we would take the point (0,-14).
$z_{min}=4^2+90=106 \qquad z_{max}=32^2+90=1114$