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Tue Nov 8 15:17:41 2011 UTC (18 months, 2 weeks ago) by aubreyja
File size: 3184 byte(s)
Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.

    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')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
14
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();