Determine all extrema for the function $f (x,y) = x^2y+y.$
hint
Posssible extremum points can only be found in the 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
Explain that the function $\,f (x,y) = x^2+4y^2-2x-4y\,$ has exactly one extremum, determine the ekstremum point and the extremum value.
hint
Possible extremum points can only be found in the stationary points of the function.
hint
Determine the Hessian matrix and find its eigenvalues? What is the meaning of the eigenvalues? Possibly check Lemma 21.17 in eNote 21.
answer
The function has one stationary point, viz. $(x,y)=(1,\frac{1}{2})$. The Hessian matrix is constant and has everywhere (and therefore also in this point) positive eigenvalues, so there is a local minimum in the point. The minimum value is $f(1,\frac 12)=-2\,$.
B
What is the difference between an extremum and a proper extremum? Is the extremum found a proper extremum?
answer
The answer to the last question is yes, see Lemma 21.17 in eNote 21.
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%Vis ved kvadratkomplettering at niveaukurven for $f$ svarende til $f(x,y)=1$ er en ellipse. Angiv ellipsens centrum og halvakser.
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%Vis at $f$ kan omskrives til en ligning på formen
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$$z-c_1=(x-c_2)^2
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### **Exercise 3: *Local Extrema for Functions of Two Variables***
Given the function f:\reel^2\rightarrow\reel by the expression $$
f(x,y)=x^3+2y^3+3xy^2-3x^2.
$$
C
Show that the points \,A=(2,0)\,, B=(1,-1)\, and \,C=(0,0)\, are stationary points for \,f\, and decide for each of these whether they are local maximum points or local minimum points. If so state the local maximum value/minimum value, and decide if they are proper.
hint
For A and B you can solve the case using the eigenvalues for the Hessian matrix in the points. C requires further investigation, e.g. you can make a sign investigation of f on the line x=0\,.
hint
Or more precisely: What happens with \,f(0,y)=2y^3\, when you pass \,y=0\,? And what does this tell you about the possibility of extremum in (0,0)?
answer
There is a proper minimum in the point \,A\, with the minimum value \,f(2,0)=-4\,. There are not ekstrema in \,B\, and \,C\,.
D
Show that the approximating second-degree polynomial for \,f\, with the development point \,A\, can be written as an equation in the unknowns \,x,y\, and \,\,z in this form:
$$</div>
z-c_3=\frac 12\,\lambda_1(x-c_1)^2+\frac 12\,\lambda_2(y-c_2)^2.
$$
Which quadratic surface does this equation describe, and what does the constants mean?
hint
Now get the equation in the form
$$</div>
z-c_3=\frac{(x-c_1)^2}{a^2}+\frac{(y-c_2)^2}{b^2}.
$$
and check the Table in Section 22.3\,.
answer
\lambda_1 and \lambda_2 are the eigenvalues for the matrix \frac 12\,\mathbf H\,.
The equation describes an upturned elliptic paraboloid with vertex \,T=(c_1,c_2,c_3)=(2,0,-4)\,. NB: The first two coordinates in \,T\, state \,A\,, while the last is the minimum value for \,f\, in \,A\,.
E
Draw the graph for f together with the graph for the approximating second-degree polynomials for f with the development points A\,, B and C\,. Discuss whether you from the eigenvalues for the Hessian matrices in the three points can decide which type of quadratic surface the second-degree polynomials describe.
### **Exercise 4: *Global Maximum and Global Minimum***
A function with the domain \reel ^2 is given by
$$
f(x,y)=xy(2-x-y)+1
$$
Let \,M\, denote the field in the \,(x,y)-plane where \,x\in\left[ 0,1\right], and y\in\left[ 0,1\right]\,.
A
Find by hand all stationary points for \,f\, on \,M\,.
answer
In \,M\, one stationary point exists, viz. \,(\frac 23,\,\frac 23)\,.
B
Determine the global maximum and minimum for \,f\, on \,M\, and the points in which these values are attained.
hint
Can we know for a fact that \,f\, has a global maximum and a global minimum on \,M\,?
hint
See the Main Theorem 21.10 in eNote 21.
hint
The global extrema are found in exception points, in the stationary points of the function or along the boundary of M, look further in Method 21.13.
hint
The investigation of the boundary is most easily done by considering the restriction of f to the relevant parts of the line segments (x,0), (0,y), (x,1) and (1,y).
hint
When you have found the stationary points and local extrema along the restrictions, you must calculate the functional values in all these points and in the endpoints of the line segments: (0,0), (1,0), (0,1) and (1,1). The point, with the largest functional value is the global maximum and the point, with the smallest value is the global minimum.
answer
Global maximum value = \,\frac{35}{27}\, attained in \,(\frac 23,\,\frac 23)\,.
Global minumium value = 1 attained on all of the line segment \,(x,0\,) for \,0\leq x \leq 1\,, on all of the line segment \,(0,y)\, for \,0\leq y \leq 1\, and in the point \,(x,y)=(1,1)\,.
C
Determine the range of \,f\, on \,M\,.
D
Plot the graph for f together with points that shows where on the graph the maximum and the minimum value are attained, and check whether your results look alright.
### **Exercise 5: *Global Maximum and Global Minimum***
Consider the function f:\reel^2\rightarrow\reel given by
$$
f(x,y)=x^2-3y^2-3xy
$$
and the set \,M=\left\{\,(x,y)\,|\,x^2+y^2\leq 1\,\right\}\,.
A
Explain that \,f\, has both a global maximum and a global minimum on \,M\, and determine these values together with the points in which they are attained.
hint
We the Main Theorem 21.10 and Method 21.13 in eNote 21
hint
The candidates to the maximum and minimum value are consituted by the stationary points and the local extrema along the boundary of \,M\,.
hint
The only stationary point for \,f\, in the domain is \,(0,0\,). Hereafter we shall investigate the restriction of \,f\, to the boundary of \,M\,.
hint
The boundary of \,M\, can be parametrized by \,(x,y)=(\cos (t),\sin(t))\, where \,t\in\left[ 0;2\pi\right]\, .
hint
The restriction of \,f\, to the boundary of \,M\, is then \,g(t)=f(\cos (t), \sin( t))\, where \,t\in\left[ 0;2\pi\right]\,. Plot the graph for \,g’(t)\,, and determine its zero points.
hint
The candidates to a global maximum and minimum are constituted by \,(0,0)\,, the points that corresponds to the solution of \,g’(t)=0\,and the value of \,g\, in the endpoints of the boundary curve (in fact there is only one endpoint, why?). Compute the functional values for \,f\, in these points. The largest functional value is then the global maximum of the function on \,M\,, and the smallest value is global minimum of the function on \,M\,.
answer
Global minimum = -\frac{7}{2} attained in (\frac{\sqrt{10}}{10},\frac{3\sqrt{10}}{10}) and (-\frac{\sqrt{10}}{10},-\frac{3\sqrt{10}}{10}).
Global maximum = \frac{3}{2} attained in (\frac{3\sqrt{10}}{10},\frac{-\sqrt{10}}{10}) and (\frac{-3\sqrt{10}}{10},\frac{\sqrt{10}}{10}).
### **Exercise 6: *Global Extrema for a Function of Three Variables***
We consider the function f:\reel^3\rightarrow \reel given by
$$
Show that \,f\, in the interior of \,\mathcal K\, only has one stationary point, viz. \,O=(0,0,0)\,, and investigate whether \,f\, has an extremum in \,O\,.
B
Determine the global maximum value and the global minimum value of \,f\, on \,\mathcal K\, and the points in which they are attained.
answer
Global max = 1 is attained in (0,1,0) and (0,-1,0)\,.
Global min = -1 is attained on the circle \left{(x,y,z)\,|\,y=0\,\,\,\,\mathrm{and}\,\,\,\,x^2+z^2=1\right}\,.
C
Determine the range of \,f\, on \,\mathcal K\,.
### **Exercise 7: *Supplementary Exercise***
Given the function f:\reel^2\rightarrow\reel with the expression $$
f(x,y)=\exp(x^2+y^2)-4xy\,
$$
A
Find all stationary points for \,f\,.
answer
The 3 stationary points are $$</div>
(0,0) \, , \, (\sqrt{\frac{1}{2} \ln{2}},\sqrt{\frac{1}{2} \ln{2}}) \, , \, (-\sqrt{\frac{1}{2} \ln{2}},-\sqrt{\frac{1}{2} \ln{2}})
$$
B
Find all extrema.
answer
There are proper minimas in the points \,(\sqrt{\frac{1}{2} \ln{2}},\sqrt{\frac{1}{2} \ln{2}}\,) \, and \, (-\sqrt{\frac{1}{2} \ln{2}},-\sqrt{\frac{1}{2} \ln{2}}\,)\, with the minimum value \,2-2 \ln 2\,.
C
Decide whether \,f\, has a global maximum or minimum, and state the values for these if they exist.
answer
There is no global maximum. Global minimum is attained in the points (\sqrt{\frac{1}{2} \ln{2}},\sqrt{\frac{1}{2} \ln{2}}\,)\, and \,(-\sqrt{\frac{1}{2} \ln{2}},-\sqrt{\frac{1}{2} \ln{2}}\,)\, with the values \,2-2 \ln 2\,.
