## DBsubject('Calculus') ## DBchapter('Differentiation in Several Variables') ## DBsection('Optimization in Several Variables') ## KEYWORDS('calculus') ## TitleText1('Calculus: Early Transcendentals') ## EditionText1('2') ## AuthorText1('Rogawski') ## Section1('14.7') ## Problem1('7') ## Author('JustAsk - Vladimir Finkelshtein') ## Institution('W.H.Freeman') DOCUMENT(); loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl"); loadMacros("Parser.pl"); loadMacros("freemanMacros.pl"); loadMacros("PGauxiliaryFunctions.pl"); loadMacros("PGgraphmacros.pl"); loadMacros("PGchoicemacros.pl"); TEXT(beginproblem()); do{ $a=non_zero_random(-4,4,1); } while($a == 2 || $a == -2);$ma = Real(-$a);$x0=non_zero_random(-8,8,1); $b=$x0*(($a)**2/2-2); Context()->texStrings;$context = Context(); $context->variables->add(y=>'Real');$f=Formula("x^2+y^2+$a*x*y+$b*x")->reduce(); $fx=Formula("2*x+$a*y+$b")->reduce();$fy=Formula("2*y+$a*x")->reduce();$y=Formula("-$a/2*x")->reduce();$ynr=Formula("-$a/2*x");$ma2=Formula("-$a/2");$y0=$y->eval(x=>$x0); $fxx=2;$fyy=2; $fxy=$a; $disc=$fxx*$fyy-($fxy)**2; if($disc>0) {$ans='a local minimum'; $wrongans1='test fails';$wrongans2='a local maximum'; $wrongans3='a saddle point';$sol="\quad f_{xx}($x0,$y0)=2>0"; $sign='>'; } if($disc<0) {$ans='a saddle point';$wrongans1='test fails'; $wrongans2='a local maximum';$wrongans3='a local minimum'; $sign='<'; }$mc = new_multiple_choice(); $mc -> qa ("","$ans",); $mc ->extra("$wrongans1","$wrongans2","$wrongans3",); BEGIN_TEXT \{ textbook_ref_exact("Rogawski ET 2e", "14.7","7") \} $PAR Find the critical point of the function $$f(x,y)=f$$$PAR $$c=$$ \{ans_rule()\} $PAR Use the Second Derivative Test to determine whether the point is \{$mc -> print_a\} $BR END_TEXT Context()->normalStrings; Context("Vector"); ANS(Point(Real($x0),Formula($y0))->cmp); Context("Numeric"); ANS(str_cmp($mc->correct_ans)); Context()->texStrings; SOLUTION(EV3(<<'END_SOLUTION')); $PAR$SOL $BR First, we find the critical point.$BR We set the first-order partial derivatives of $$f(x,y)=f$$ equal to zero and solve $f_x(x,y)=fx=0$ $f_y(x,y)=fy=0$ The second equation implies that $$y=ynr$$. Substituting into the first and solving for $$x$$ gives: $2x+ynr+b=0 \Rightarrow x=x0$ The corresponding value of $$y$$ is $$y=ma2(x0)=y0$$. The critical point is $$(x0,y0)$$. $PAR Now we compute the Discriminant. We find the second-order partials: $f_{xx}(x,y)=2 \quad f_{yy}(x,y)=2 \quad f_{xy}(x,y)=a$ The discriminant is $D(x,y)=f_{xx}f_{yy}-f_{xy}^2=2\cdot 2-(a)^2=disc$$PAR Applying the Second Derivative Test we have $D(x0,y0)=disc sign 0 sol$ The Second Derivative test implies that $$f(x0,y0)$$ is $ans.$BR END_SOLUTION ENDDOCUMENT();