Chapter 4 111

# Chapter 4 111 - APPLIED PROJECT THE SHAPE OF A CAN 373(c...

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Unformatted text preview: APPLIED PROJECT THE SHAPE OF A CAN 373 (c) Substituting d = 10 into the equations for 1(w) and I'(m) gives 110(32) : W100 + m and 2x 22(1‘ — 10) Ii0(\$): ———2 ‘ '—“———“‘_2 (2:2 + 100)2 v 20cc + 200)2 0017 00006 From the graphs. it seems that for d = 10. the intensity is minimized at the endpoints. that is. as = 0 and m = 10. The midpoint is now the most brightly lit 0014 70. 00-06 pomt! (d) From the ﬁrst ﬁgures in parts (b) and (c). we see that the minimal illumination changes from the midpoint (ac : 5 with d = 5) to the endpoints (an : 0 and m = 10 with d : 10). 0 0365 0 023 0.01 0-10 0.1005 w”) 000325 0.0215 —00-l So we try d : 6 (see the ﬁrst ﬁgure) and we see that the minimum value still occurs at a; = 5. Next. we let d : 8 (see the second ﬁgure) and we see that the minimum value occurs at the endpoints. It appears that for some value of d between 6 and 8. we must have minima at both the midpoint and the endpoints. that is. 1(5) must equal [(0). To ﬁnd this value of d. we solve [(0) : [(5) (with k : 1): i+_1_, 1 + 1 _ 2 :> d2 100+d2_25+d2 25+d2725+d2 (25+d2)(100+d2) +d2(25+d2) = 2d2(100+d2) => 2500 + 125012 + d4 + 25d2 + d4 — 200d2 : 2d4 > 2500 50d2 » d2 50 d = 5 \/§ % 7.071 (for O S d g 10). The third ﬁgure. a graph of [(0) — [(5) with d independent. conﬁrms that [(0) ~ [(5) = 0, that is, [(0) = 1(5), when d : 5 ﬂ. Thus. the point of minimal illumination changes abruptly from the midpoint to the endpoints when d : 5 x/i APPLIED PROJECT The Shape of a Can 1. In this case. the amount of metal used in the making of each top or bottom is (2r)2 : 4r? So the quantity we want to minimize is A : 27rrh + 2(4r2). But V : 7TT2h 4:) h : V/7rr2. Substituting this expression for h in A gives A : 2V/r + 8T2. Differentiating A with respect to 7". we get dA/dr = —2V/7°2 + 167" : 0 => d2A 4V = 1 _ CW 6 + T3 > 0 167‘3 : 2V : 27r7'2h <1} =lloo — m 2.55. This gives a minimum because r ...
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