Determine all extrema of the function $f (x,y) = x^2y+y.$
hint
Possible extremum points can be found at stationary points of the function.
answer
The function has no stationary points. Therefore no extrema.
Exercise 2: Application of the Hessian Matrix
This exercise should be solved by hand.
A
Show that the function $\,f (x,y) = x^2+4y^2-2x-4y\,$ has exactly one extremum. Determine the extremum point, investigate whether it is a maximum or a minimum, and compute the extremum value.
hint
Possible extremum points can only be found at stationary points of the function.
hint
Determine the Hessian matrix and find its eigenvalues. See Lemma 21.17 in eNote 21.
answer
The function has one stationary point:
$$(x,y)=\left(1,\frac{1}{2}\right).$$
The Hessian matrix is constant and has everywhere (and therefore also at this point) positive eigenvalues, so there is a local minimum at the point. The minimum value is
$$f\left(1,\frac 12\right)=-2\,.$$
B
What is the difference between an extremum and a proper extremum? Is the result above a proper extremum?
answer
The answer to the last question is yes, see Lemma 21.17 in eNote 21.
Exercise 3: Local Extrema for Functions of Two Variables
We are given the function $f:\reel^2\rightarrow\reel$ with the expression
$$f(x,y)=x^3+2y^3+3xy^2-3x^2.$$
A
Show that the points $\,A=(2,0)\,,$$B=(1,-1)\,$ and $\,C=(0,0)\,$ are stationary points of $\,f\,$ and investigate for each of these whether they are local maximum points or local minimum points. If so, state the local maximum value/minimum value, and decide whether they are proper.
hint
Investigate with the eigenvalues of the Hessian matrix at the points. You will find that point $C$ requires further investigation - attempt a sign investigation on $f$ along the line $x=0\,.$
hint
What happens with $\,f(0,y)=2y^3\,$ when you pass $\,y=0\,$? And what does this tell you about the possibility of extremum at $(0,0)$?
hint
Use the double-prime criterion.
answer
There is a proper minimum at $\,A\,$ with the minimum value $\,f(2,0)=-4\,.$ There is not extrema at $\,B\,$ nor $\,C\,.$
B
Show that the approximating second-degree polynomial of $\,f\,$ with development point $\,A\,$ can be written as an equation with the unknowns $x,y$ and $z$ on this form:
$\lambda_1$ and $\lambda_2$ are the eigenvalues of the matrix $\frac 12\,\mathbf H\,$.
The equation describes an “upwards-turned” elliptic paraboloid with vertex $\,T=(c_1,c_2,c_3)=(2,0,-4)\,.$
NB: The first two coordinates of $\,T\,$ are identical to $\,A\,,$ and the last coordinate is the minimum value of $\,f\,$ at $\,A\,.$
C
Draw the graph of $f$ together with the graph of the approximating second-degree polynomials of $f$ with the development points $A\,,$$B$ and $C\,.$ Discuss whether you from the eigenvalues of the Hessian matrices at these three points can decide which type of quadratic surface the second-degree polynomials describe.
Exercise 4: Global Maximum and Global Minimum
A function with domain $\reel ^2$ is given by
$$f(x,y)=xy(2-x-y)+1.$$
Let $\,M\,$ denote the region in the $\,(x,y)$-plane where $\,x\in\left[ 0,1\right]$ and $y\in\left[ 0,1\right]\,.$
A
Find by hand all stationary points of $\,f\,$ on $\,M\,.$
answer
There is one stationary point of $f$ on $\,M\,$:
$$\left(\frac 23,\,\frac 23\right).$$
B
Determine the global maximum and global minimum of $\,f\,$ on $\,M\,$ along with the points at which these values are attained.
hint
Can we know for a fact that $\,f\,$ has a global maximum and a global minimum on $\,M\,?$
hint
See the Main Theorem 21.10 in eNote 21.
hint
The global extrema are found at exceptional points, at stationary points of the function or at the boundary of $M$. See Method 21.13.
hint
The investigation of the boundary is most easily done by considering the restriction of $f$ on the relevant parts of the line segments $(x,0)$, $(0,y)$, $(x,1)$ and $(1,y).$
hint
When you have found the stationary points and local extrema along the restrictions, you must calculate the function values at all these points and at the endpoints of the line segments: $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1).$ The point, with the largest function value is the global maximum, and the point with the smallest value is the global minimum.
answer
Global maximum is found to be $\,\frac{35}{27}\,,$ attained at $\,(\frac 23,\,\frac 23)\,.$
Global minimum is found to be $1\,,$ attained on the entire line segment $\,(x,0\,)$ for $\,0\leq x \leq 1\,$, on the entire line segment $\,(0,y)\,$ for $\,0\leq y \leq 1\,,$ as well as at the point $\,(x,y)=(1,1)\,.$
C
Determine the range of $\,f\,$ on $\,M\,.$
answer
$$f(M)=\left[1\,,\,\frac{35}{27}\right]$$
D
Plot the graph of $f$ with points that show where on the graph the maximum and the minimum value are attained, and do a visual check of whether your results look alright.
Exercise 5: Global Maximum and Global Minimum
Consider the function $f:\reel^2\rightarrow\reel$ given by
$$f(x,y)=x^2-3y^2-3xy$$
and the set $\,M=\lbrace\,(x,y)\,|\,x^2+y^2\leq 1\,\rbrace\,.$
A
Explain that $\,f\,$ has both a global maximum and a global minimum on $\,M\,.$ Determine these values and state the points at which they are attained.
hint
See the Main Theorem 21.10 and Method 21.13 in eNote 21.
hint
Candidates for the maximum and minimum values are comprised by stationary points in the interior and local extrema on the boundary of $\,M\,$.
hint
The only stationary point of $\,f\,$ in the domain is $\,(0,0\,)$. Now do an investigation of the restriction of $\,f\,$ to the boundary of $\,M\,.$
The restriction of $\,f\,$ to the boundary of $\,M\,$ is then $\,g(t)=f(\cos (t), \sin( t))\,$ where $\,t\in\left[ 0;2\pi\right]\,$. Plot the graph of $\,g’(t)\,$, and determine its roots (the points at which the function is zero).
hint
The candidates for global maximum and global minimum are comprised by $\,(0,0)$, the points that correspond to the solution to $\,g’(t)=0\,,$ and the value of $\,g\,$ at the end points of the boundary curve (in fact there is only one end point – why?). Compute the function values of $\,f\,$ at all these points. The largest function value is then the global maximum of the function on $\,M\,$, and the smallest value the global minimum.
answer
Global minimum is found to be $-\frac{7}{2}$, attained at $\left(\frac{\sqrt{10}}{10},\frac{3\sqrt{10}}{10}\right)$ and $\left(-\frac{\sqrt{10}}{10},-\frac{3\sqrt{10}}{10}\right)$.
Global maximum is foudn to be $\frac{3}{2}$, attained at $\left(\frac{3\sqrt{10}}{10},\frac{-\sqrt{10}}{10}\right)$ and $\left(\frac{-3\sqrt{10}}{10},\frac{\sqrt{10}}{10}\right)$.
Exercise 6: Global Extrema for a Function of Three Variables
We consider a function $f:\reel^3\rightarrow \reel$ given by
Show that $\,f\,$ in the interior of $\,\mathcal K\,$ only has one stationary point, that being $\,O=(0,0,0)\,,$ and investigate whether $\,f\,$ has an extremum at $\,O\,.$
hint
The approach with equating the gradient to the zero vector and solving as well as the method of investigating Hessian matrix eigenvalues applies to functions of any number of variables, not just two.
hint
Solve $\nabla f(x,y,z)=(0,0,0)$ to find $O$. Then compute the eigenvalues of $H(O)$.
B
State the global maximum value and the global minimum value of $\,f\,$ on $\,\mathcal K\,$ and the points at which these values are attained.
answer
Global maximum is $1$, attained at $(0,1,0)$ and $(0,-1,0)\,$.
Global minimum is $-1$, attained on the circle $\lbrace(x,y,z)\,|\,y=0\,\,\,\,\mathrm{and}\,\,\,\,x^2+z^2=1\rbrace\,$.
C
Determine the range of $\,f\,$ on $\,\mathcal K\,.$
answer
$$f(\mathcal K)=[-1,1]$$
Exercise 7: Supplementary Exercise
We are given a function $f:\reel^2\rightarrow\reel$ with the expression
There are proper minima at the points $\,\left(\sqrt{\frac{1}{2} \ln{2}}\,,\,\sqrt{\frac{1}{2} \ln{2}}\,\right) \,$ and $\, \left(-\sqrt{\frac{1}{2} \ln{2}}\,,\,-\sqrt{\frac{1}{2} \ln{2}}\,\right)\,$ with the minimum value $\,2-2 \ln 2\,.$
C
Decide whether $\,f\,$ has a global maximum or minimum, and state the values of these if they exist.
answer
There is no global maximum. Global minimum is attained at the points $\left(\sqrt{\frac{1}{2} \ln{2}}\,,\,\sqrt{\frac{1}{2} \ln{2}}\,\right)\,$ and $\,\left(-\sqrt{\frac{1}{2} \ln{2}}\,,\,-\sqrt{\frac{1}{2} \ln{2}}\,\right)\,$ with the minimum value $\,2-2 \ln 2\,.$
