Find the absolute maximum and minimum

Question

Find the absolute minimum and maximum of f(x, y) = 2x² - y² + 6y over the 2 region x² + y² ≤ 16. You must look for local maximums and minimums in the region as well as check along the boundary.

Answer 1

The gradient is

(4x, 6 - 2y), which has a 0 at (0, 3), which is in the interior. However, the second derivative is

4 0

0 -2

which tells us that this is a saddle point as eigenvalues are positive (4) and negative (-2)

We also can see that the minimum will be at (0, -4) and is 2(0)² - (-4)² +6(-4) = -40

However, we still will have to use the standard LaGrange procedure for the maximum. (This will also show us that the stated minimum is one of the candidates, s ptoog og yr

l(x,y) = 2x² - y² + 6y + l(x²+y² - 16)

The gradient is (4x + 2xl, -2y + 6 + 2yl) = (2x(2 + l), -2y + 6 + 2yl)

We get the minimum mentioned above by setting 2x(2 + l) = 0 by letting x = 0, which leaves us free to vary l such that we get -2y + 6 + 2yl = 0 and are on the boundary. We can just plug in y = 4 and -4.

f(0, -4) is given above. f(0, 4) = 8. This will be a candidate for the minimum and maximum. Clearly, it is not the minimum. We could look at l if we chose, but that is not necessary.

2x(2 + l) = 0, also, when l = -2

Then, -2y + 6 + 2yl = 0

-2y + 6 + 2y(-2) = 0

-6y + 6 = 0

y = 1

As x² + y² = 16, x = ±√15.

f(-√15, 1) = f(√15, 1) = 35

This will be a maximum. (The other value at (0, 4) now makes sense, as within the boundary, this becomes a local minimum).

The absolute minimum is -40 at (0, -4)

The absolute maximum is 35 at (-√15, 1) and (√15, 1)