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Exam2_S2009 Solutions

Now we must solve the system f x y 4 λx 0 1 f y x 6

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Now we must solve the system F x = y + 4 λx = 0 (1) F y = x + 6 λy = 0 (2) F λ = 2 x 2 + 3 y 2 9 = 0 . (3) Multiplying (1) by 3 y and (2) by 2 x , we get the two equations F x = 3 y 2 + 12 λxy = 0 (4) F y = 2 x 2 12 λxy = 0 . (5) Adding (4) and (5), we get the new equation 3 y 2 2 x 2 = 0 3 y 2 = 2 x 2 . (6) Substituting (6) into (3), we have 3 y 2 + 3 y 2 9 = 0 6 y 2 = 9 y = ± radicalbigg 3 2 . If y = radicalbig 3 / 2 then (6) gives x = ± 3 2 . Likewise, if y = radicalbig 3 / 2 then (6) gives x = ± 3 2 . Thus we have four critical points: parenleftBigg 3 2 , radicalbigg 3 2 parenrightBigg , parenleftBigg 3 2 , radicalbigg 3 2 parenrightBigg , parenleftBigg 3 2 , radicalbigg 3 2 parenrightBigg , parenleftBigg 3 2 , radicalbigg 3 2 parenrightBigg . The function values for each of the critical points above are, respectively, 3 2 radicalbigg 3 2 , 3 2 radicalbigg 3 2 , 3 2 radicalbigg 3 2 , 3 2 radicalbigg 3 2 . Thus 3 / 2 radicalbig 3 / 2 is our maximum and 3 / 2 radicalbig 3 / 2 is our minimum. 6
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9. You appreciate your calculus instructor so much that you wish to set up a savings account from which he/she can withdraw $2,000 per month. Supposing that the savings account earns 10%/year compounded continuously and that your calculus instructor lives forever, how much money would you need to put into that account right now to make your wish come true. Since we wish to withdraw money from the account forever, this is an example of a perpetuity. We want $2,000 per month, so P = 2000 and m = 12. The interest rate is r = 0 . 1. Thus the present value of this perpetuity is V = (12)(2000) 0 . 1 = $240 , 000 . 10. Evaluate the following improper integrals whenever they are convergent. a. integraldisplay 1 1 x + 2 dx integraldisplay 1 1 x + 2 dx = lim b →∞ integraldisplay b 1 1 x + 2 dx ( u -substitution with u = x + 2) = lim b →∞ ln | x + 2 | vextendsingle vextendsingle vextendsingle b 1 = lim b →∞ ln | b + 2 | − ln | 1 + 2 | This limit diverges since ln( b ) goes to infinity when b goes to infinity.
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