View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/14_Differentiation_in_Several_Variables/14.7_Optimization_in_Several_Variables/14.7.21.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Differentiation in Several Variables')
    3 ## DBsection('Optimization in Several Variables')
    4 ## KEYWORDS('calculus')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('14.7')
    9 ## Problem1('21')
   10 ## Author('JustAsk - Vladimir Finkelshtein')
   11 ## Institution('W.H.Freeman')
   12 
   13 DOCUMENT();
   14 
   15 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   16 loadMacros("Parser.pl");
   17 loadMacros("freemanMacros.pl");
   18 loadMacros("PGauxiliaryFunctions.pl");
   19 loadMacros("PGgraphmacros.pl");
   20 loadMacros("PGchoicemacros.pl");
   21 
   22 TEXT(beginproblem());
   23 
   24 $y0=non_zero_random(-3.5,3.5,0.5);
   25 $b=non_zero_random(1,6,1);
   26 $a=-2*$y0*$b;
   27 $xnum=-$y0*$a+1;
   28 $x="\frac{$xnum}{$a}";
   29 $x0=(-$y0*$a+1)/$a;
   30 
   31 Context()->texStrings;
   32 $context = Context();
   33 $context->variables->add(y=>'Real');
   34 
   35 
   36 #$f=Formula("$a*x-$b*y^2-ln(x+y)")->reduce();
   37 $f=Formula("$a*x-$b*y^2-ln|x+y|")->reduce();
   38 $fx=Formula("$a-1/(x+y)")->reduce();
   39 $fy=Formula("-2*$b*y-1/(x+y)")->reduce();
   40 
   41 $f2=Formula("-2*$b*y-$a")->reduce();
   42 
   43 $fxx=Formula("1/(x+y)^2")->reduce();
   44 $fyy=Formula("-2*$b+1/(x+y)^2")->reduce();
   45 $fxy=Formula("1/(x+y)^2")->reduce();
   46 
   47 $disc=Formula("-2*$b/(x+y)^2")->reduce();
   48 $d0=$disc->eval(x=>$x0,y=>$y0);
   49 
   50 $mc = new_multiple_choice();
   51 $mc -> qa ("","a saddle point",);
   52 $mc ->extra("a local minimum","a local maximum","test fails",);
   53 
   54 BEGIN_TEXT
   55 \{ textbook_ref_exact("Rogawski ET 2e", "14.7","21") \}
   56 $PAR
   57 Find the critical point of the function \(f(x,y)=$f\).
   58 $PAR
   59 \(c=\) \{ans_rule()\}
   60 $BR
   61 Use the Second Derivative Test to determine whether it is
   62 \{$mc -> print_a\} $PAR
   63 $BR
   64 END_TEXT
   65 
   66 Context()->normalStrings;
   67 Context("Point");
   68 ANS(Point(Real($x0),Real($y0))->cmp);
   69 Context("Numeric");
   70 ANS(str_cmp($mc->correct_ans));
   71 Context()->texStrings;
   72 
   73 SOLUTION(EV3(<<'END_SOLUTION'));
   74 $PAR
   75 $SOL
   76 $BR
   77 First, we find the critical point.$BR
   78 We set the first-order partial derivatives of \(f(x,y)=$f\) equal to zero and solve
   79 \[f_x(x,y)=$fx=0\]
   80 \[f_y(x,y)=$fy=0\]
   81 The first equation implies that \(\frac{1}{x+y}=$a\). Substituting into the second equation gives
   82 \[$f2=0 \quad \Rightarrow \quad y=$y0\]
   83 We substitute \(y=$y0\) in the first equation and solve for \(x\):
   84 \[$a-\frac{1}{x+$y0}=0 \quad \Rightarrow \quad x+$y0=\frac{1}{$a} \quad \Rightarrow \quad x=\frac{-$y0 \cdot $a + 1}{$a}=$x \]
   85 We obtain critical point \(\left($x, $y0\right)\). $BR
   86 Notice, that although \(f_x\) and \(f_y\) do not exist where \(x+y=0\), these are not critical points, since \(f\) is not defined at these points.
   87 $PAR
   88 Now we compute the Discriminant. We find the second-order partials:
   89 \[f_{xx}(x,y)=\frac{\partial}{\partial x}\left($fx\right)=$fxx\]
   90 \[f_{yy}(x,y)=\frac{\partial}{\partial y}\left($fy\right)=$fyy\]
   91 \[f_{xy}(x,y)=\frac{\partial}{\partial y}\left($fx\right)=$fxy\]
   92 The discriminant is thus
   93 \[D(x,y)=f_{xx}f_{yy}-f_{xy}^2=$fxx \cdot \left($fyy\right)-\frac{1}{(x+y)^4}=$disc\]
   94 $PAR
   95 Applying the Second Derivative Test we have
   96 \[D\left($x,$y0\right)=$d0<0\]
   97 We conclude that \(f\left($x,$y0\right)\) is a saddle point.
   98 $BR
   99 END_SOLUTION
  100 
  101 ENDDOCUMENT();