In the right triangle ABC with right angle at A , the inradius is 2cm. If the hypothenuse BC=10cm , determine the perimeter and
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Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
Taken together, we find that (0, 0, 0) appears to be the only critical point on
within the ball. At this point, we have 
.
Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian
with partial derivatives (set to 0)
We then observe that
So, ignoring the critical point we've already found at (0, 0, 0),
So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of
, at which points we get a value of either of 
, with the maximum being the positive value and the minimum being the negative one.
Taken together, we find that (0, 0, 0) appears to be the only critical point on
Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian
with partial derivatives (set to 0)
We then observe that
So, ignoring the critical point we've already found at (0, 0, 0),
So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of