MAM1004H TUTORIAL SOLUTIONS 10

Mathematics MAM1004HTUTORIAL 10 Selected Answers 20061. Suppose we cut the wire x metres from one end. We turnthe x-metre length into a square. Since the square has allfour sides equal and their lengths must add up to x, wexsee that each side has length . Thus our square has area42x x x£ = .4 4 16The rest of the wire, 10¡x metres long, gets turned into acircle. So the 10¡x metres goes into making the circum-ferenceofthiscircle. Iftheradiusofthecircleis r, then we10¡xknow that 2…r =10¡x, so that gives us r = . Now we2…2 2(10¡x) (10¡x)2canflndtheareaofthecircle: itis…r =… = .2(2…) 4…Now we have a formula for the total area enclosed by thesetwo shapes; it is2 2x (10¡x)A(x)= + :16 4…Notice that the values for x that make sense lie between 00and 10. Let’s draw a sign table for A(x) on the interval[0;10]andthenflndtheglobalmaximumandminimumforA(x) on that interval.2x 2(10¡x)(¡1) x x¡100A(x)= + = + :16 4… 8 2…40Setting this equal to zero yields one solution: x = ……+405:601. Note that A(x) is never undeflned.1x 0 5:601 100A(x) ¡ 0 +A(x) decr local min incr100Looking at endpoints: A(0) = … 7:9577. (That’s the4…area if the square gets no area at all.)210At the other endpoint, A(10) = = 6:25. (That’s how16much area you get if there is no circle).At 5:601, we have A(5:601) … 3:5006, so we see that theglobal minimum occurs here. The global maximum occurswhen x = 0, which is really telling us that we ought toforget about the square all together and ...