What are the values and types of the critical points, if any, of #f(x,y)=4y(1 - x^2)#?

Question

Evelyn Agard · Accepted Answer

There are two critical points, with coordinates #(x,y)=(pm 1, 0)#, which are both saddle points.

Let #z=f(x,y)=4y(1-x^2)=4y-4x^2y#, then #(partial z)/(partial x)=-8xy# and #(partial z)/(partial y)=4-4x^2#. After setting both of these partial derivatives equal to zero, we get that #4=4x^2# so that #x^2=1# and #x= pm 1# and we also have #-8xy=0#, so that #y=0#. The two critical points are therefore #(x,y)=(pm 1, 0)#. The second-order partial derivatives are #(partial^{2} z)/(partial x^{2})=-8y#, #(partial^{2} z)/(partial y^{2})=0#, and #(partial^{2} z)/(partial x partial y)=-8x#, giving a discriminant of #(partial^{2} z)/(partial x^{2}) * (partial^{2} z)/(partial y^{2})-((partial^{2} z)/(partial x partial y))^{2}=0-(-8x)^{2}=-64x^{2}# This discriminant is negative at both critical points, making them both saddle points.

Evelyn Agard · Answer

undefined The critical points of the function $ f(x, y) = 4y(1 - x^2) $ occur where both partial derivatives, $ \frac{\partial f}{\partial x} $ and $ \frac{\partial f}{\partial y} $, are equal to zero. Calculating these partial derivatives, we find:

$$ \frac{\partial f}{\partial x} = -8xy $$
$$ \frac{\partial f}{\partial y} = 4(1 - x^2) $$

Setting both equal to zero gives us:

$$ -8xy = 0 \Rightarrow x = 0 \text{ or } y = 0 $$
$$ 4(1 - x^2) = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 $$

Thus, the critical points are $ (0, 0) $, $ (1, 0) $, and $ (-1, 0) $.

To determine the type of each critical point, we can use the second partial derivative test. Calculating the second partial derivatives, we get:

$$ \frac{\partial^2 f}{\partial x^2} = -8y $$
$$ \frac{\partial^2 f}{\partial y^2} = 0 $$
$$ \frac{\partial^2 f}{\partial x \partial y} = -8x $$

The determinant of the Hessian matrix, $ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $, at each critical point is:

For $ (0, 0) $:
$$ D = -8 \cdot 0 - (-8 \cdot 0)^2 = 0 $$
Since $ D = 0 $, the test is inconclusive.

For $ (1, 0) $:
$$ D = -8 \cdot 0 - (-8 \cdot 1)^2 = -64 $$
Since $ D < 0 $, this critical point is a saddle point.

For $ (-1, 0) $:
$$ D = -8 \cdot 0 - (-8 \cdot -1)^2 = -64 $$
Since $ D < 0 $, this critical point is also a saddle point.