in this function (if exist); Find the maximum-minimum and saddle points.

Question

f(x,y)= 2x³+xy²+5x²+y²

Answer 1

Hi Aktug,

Any max/min/saddle points must occur at a critical point, which is a point where both partial derivatives are equal to 0 (in general, we also need to consider points where at least one of the partial derivatives fails to exist, but since your function is a polynomial the partial derivatives exist everywhere).

The partial derivatives are f_x = y² + 6x² + 10x and f_y = 2y + 2xy. To find the critical points, we therefore need to find all possible solutions to the following system of equations:

y² + 6x² + 10x = 0 (1)

2y + 2xy = 0 (2)

The second equation factorizes as 2y(1+x) = 0, so this holds if y = 0 or x = -1. We now separately plug these into the first equation.

If y = 0, then the first equation becomes 6x² + 10x = 0 or 2x(3x + 5) = 0, which holds if x = 0 or x = - 5/3. This means that (x,y) = (0,0), (-3/5, 0) are critical points.

Now suppose x = -1. Then the first equation becomes

y² + 6 - 10 = 0

y² - 4 = 0

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

which has solution y = -2 or y = 2. This means that (x,y) = (-1,-2), (-1,2) are also critical points.

We now need to classify each of the critical points (x,y) = (0,0), (-3/5, 0), (-1,-2), (-1,2) that we found above.

This is done using the second partial derivatives as follows. First compute the function

D = f_xx f_yy - (f_xy)²

and evaluate it at each critical point. Then

1. If D > 0 and f_xx > 0, the critical point is a relative min.
2. If D > 0 and f_xx < 0, the critical point is a relative max.
3. If D < 0, then the critical point is a saddle point.
4. If D = 0, the test is indeterminate, and we need to use other techniques to classify the critical point.

Since the second partial derivatives are f_xx = 12x + 10, f_yy = 2x + 2, f_xy = 2y, the function D is given by

D = (12x + 10)(2x + 2)-(2y)².

Let's now investigate each critical point:

1. At (x,y) = (0,0), D = 20>0 and f_xx = 10>0, so (0,0) is a relative min.
2. At (x,y) = (-3/5,0), D = 56/25 >0 and f_xx = 14/5>0, so (-3/5,0) is a relative min.
3. At (x,y) = (-1,-2), D = -16 < 0, so (-1,-2) is a saddle point.
4. At (x,y) = (-1,2), D = -16 < 0, so (-1,2) is a saddle point.

I hope this helps!