Worked examples · Sheet 2

Critical points of functions with exponentials and logarithms

Theory used: Critical point Classification with the Hessian matrix

Exercise 1

Find and classify the critical points of the function

f (x, y) = (x^{2} + y^{2}) e^{- x} .

{f_{x} = (2 x - x^{2} - y^{2}) e^{- x} = 0 f_{y} = 2 y e^{- x} = 0 ⟹ y = 0, 2 x - x^{2} = 0 ⟹ (0, 0), (2, 0)

f_{xx} = (x^{2} + y^{2} - 4 x + 2) e^{- x}, f_{y y} = 2 e^{- x}, f_{x y} = - 2 y e^{- x}

H_{f} (0, 0) H_{f} (2, 0) = (2002), = (- 2 e^{- 2} 0 0 2 e^{- 2}), det H det H = 4 > 0, f_{xx} = 2 > 0 = - 4 e^{- 4} < 0

f (0, 0) = 0, f (2, 0) = 4 e^{- 2}

The factor $e^{- x}$ is positive for every $x$ , hence the system reduces to setting the remaining factors to zero: from the second equation it follows that $y = 0$ , and the first becomes $2 x - x^{2} = x (2 - x) = 0$ , from which $x = 0$ or $x = 2$ . To compute $f_{xx}$ we differentiate the product $(2 x - x^{2} - y^{2}) e^{- x}$ with respect to $x$ and factor out $e^{- x}$ . At the origin the Hessian is positive definite, therefore $(0, 0)$ is a local minimum point; the minimum is also absolute, since $f (x, y) \geq 0 = f (0, 0)$ for every $(x, y)$ , as $f$ is the product of two nonnegative quantities. At $(2, 0)$ the determinant is negative, therefore the point is a saddle point.

Exercise 2

Find and classify the critical points of the function

f (x, y) = x y e^{- x - y} .

{f_{x} = y (1 - x) e^{- x - y} = 0 f_{y} = x (1 - y) e^{- x - y} = 0 ⟹ (0, 0), (1, 1)

f_{xx} = - y (2 - x) e^{- x - y}, f_{y y} = - x (2 - y) e^{- x - y}, f_{x y} = (1 - x) (1 - y) e^{- x - y}

H_{f} (0, 0) H_{f} (1, 1) = (0110), = (- e^{- 2} 0 0 - e^{- 2}), det H det H = - 1 < 0 = e^{- 4} > 0, f_{xx} = - e^{- 2} < 0

f (1, 1) = e^{- 2}

Since $e^{- x - y} > 0$ , the system is equivalent to $y (1 - x) = 0$ and $x (1 - y) = 0$ and is solved by cases. If $y = 0$ , the second equation becomes $x = 0$ and we obtain the origin; if $x = 1$ , the second equation becomes $1 - y = 0$ , that is $y = 1$ , and we obtain $(1, 1)$ . At the origin the determinant of the Hessian is negative, therefore the point is a saddle point; on the two axes the function is zero, while in the first and third quadrants it is positive and in the second and fourth it is negative, in agreement with the classification. At $(1, 1)$ the determinant is positive and $f_{xx} < 0$ , therefore the point is a local maximum point with $f (1, 1) = e^{- 2}$ . The maximum is not absolute, since along the bisector $y = x$ , as $x \to - \infty$ , we have $f (x, x) = x^{2} e^{- 2 x} \to + \infty$ .

Exercise 3

Find and classify the critical points of the function

f (x, y) = x^{2} + y^{2} - ln (x y) .

domain: x y > 0

⎩ ⎨ ⎧ f_{x} = 2 x - \frac{1}{x} = 0 f_{y} = 2 y - \frac{1}{y} = 0 ⟹ x = \pm \frac{2}{2}, y = \pm \frac{2}{2}

P_{1} = (\frac{2}{2}, \frac{2}{2}), P_{2} = (- \frac{2}{2}, - \frac{2}{2})

H_{f} (x, y) = 2 + \frac{1}{x ^{2}} 0 0 2 + \frac{1}{y ^{2}}, H_{f} (P_{1}) = H_{f} (P_{2}) = (4004)

f (P_{1}) = f (P_{2}) = \frac{1}{2} + \frac{1}{2} - ln \frac{1}{2} = 1 + ln 2

The domain is the set $x y > 0$ , that is, the union of the open first and third quadrants: every critical point must be sought there. The equation $2 x - \frac{1}{x} = 0$ is equivalent to $x^{2} = \frac{1}{2}$ , hence $x = \pm \frac{2}{2}$ , and the same holds for $y$ . The combinations with opposite signs do not belong to the domain, therefore the critical points are only $P_{1}$ and $P_{2}$ . At both of them the Hessian is the diagonal matrix with entries $4$ and $4$ , which is positive definite, hence both are local minimum points with value $1 + ln 2$ ; the two values coincide because $f (- x, - y) = f (x, y)$ . In each of the two quadrants the minimum is absolute, since $f \to + \infty$ both toward the boundary of the quadrant, where $ln (x y) \to - \infty$ , and as $x^{2} + y^{2} \to + \infty$ .