What are the values and types of the critical points, if any, of #f(x,y) = sqrty(5-2x^2-3y ))#?

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Emily Ammerman · Accepted Answer

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Cameron Alexander · Answer

undefined To find the critical points of $ f(x,y) = \sqrt{5 - 2x^2 - 3y} $, we first need to find the partial derivatives with respect to $ x $ and $ y $ and then solve for where they are simultaneously equal to zero.

Taking the partial derivatives:

$$ \frac{\partial f}{\partial x} = \frac{-4x}{\sqrt{5 - 2x^2 - 3y}} $$
$$ \frac{\partial f}{\partial y} = \frac{-3}{2\sqrt{5 - 2x^2 - 3y}} $$

Setting both partial derivatives equal to zero:

$$ \frac{-4x}{\sqrt{5 - 2x^2 - 3y}} = 0 $$
$$ \frac{-3}{2\sqrt{5 - 2x^2 - 3y}} = 0 $$

Solving these equations yields:

$$ x = 0 $$
$$ y = \frac{5}{3} $$

To determine the nature of the critical point, we need to evaluate the second partial derivatives and the discriminant of the Hessian matrix.

The second partial derivatives are:

$$ \frac{\partial^2 f}{\partial x^2} = \frac{-4(5 - 2x^2 - 3y) + 8x^2}{2(5 - 2x^2 - 3y)^{3/2}} $$
$$ \frac{\partial^2 f}{\partial y^2} = \frac{9}{4(5 - 2x^2 - 3y)^{3/2}} $$
$$ \frac{\partial^2 f}{\partial x \partial y} = 0 $$

Evaluating these at the critical point $ (0, 5/3) $ gives:

$$ \frac{\partial^2 f}{\partial x^2} = \frac{10}{9\sqrt{5}} > 0 $$
$$ \frac{\partial^2 f}{\partial y^2} = \frac{9}{10\sqrt{5}} > 0 $$

Since both second partial derivatives are positive, the Hessian matrix is positive definite, indicating that the critical point is a local minimum. Therefore, the critical point $ (0, 5/3) $ is a minimum point.