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Chapter 4 99 - SECTION 4.7 0PTlMlZATlON PROBLEMS 361 a%r or...

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Unformatted text preview: SECTION 4.7 0PTlMlZATlON PROBLEMS 361 a: : %r or a: : ~r. Now A(r) = 0 = A(—r) 2 the maximum occurs where m = ér. so the triangle has height r-I- —;—r = grandbaseh/r2 — (%r)2 : 2‘/%7'2 2 \/3T. , . , 3 — y 3 24. The rectangle has area my. By Slmllal‘ triangles w = 1 => —4y +12 2 3m ory : aim + 3. So the area is 14(30): w(—%cc + 3) = —%m2 + 31: where-0 S :c g 4. Now I ‘ A _ 3 ' 0—A($)i gmi3 m 2andy §.Since A(O) : [1(4) : 0, the maximum area is A(2) = 2(%) : 3 cm2. 25. 6 The cylinder has volume V : 7ry2(2m). Also 302 + y2 : r2 => y2 : r2 , 232, so V(:c) = 7T(r2 — 332)(2a:) : 27r(r2:c — :03). where 0 g m g r. V’(a:) : 27r(7"2 — 3:32) : 0 :> :c : r/\/§. Now k V(0) : V(7~) : 0. so there is a maximum when a: : r/\/§ and V V(r/\/3) =7r(r2—r2/3)(2r/\/3) :47rr3/(3x/3). 26. By similar triangles, y/J: : h/r. so y : hm/r. The volume of the AA cylinder is 7rm2(h i y) : 7rth 7 (7rh/r)m3 : V($). Now ‘lh h 2 ' V/(m) : 27rhzr — (37rh/r)a: : 7rh$(2 * 3x/7’). 3 J So V'($) : 0 :> a: = 0 or :c = 27". The maximum clearly occurs when a: : gr and then the volume is r Trim2 — (Trh/r)a:3 = 7rhm2(1 — 17/1“): 7r(§'r)2 h(1— 2) = AFT2h. 27 27. 6 The cylinder has surface area 2(area of the base) + (lateral surface area) = 27r(radius)2 + 27r(radius)(height) : 27ry2 + 27ry(2$). k Nowzc2+y22r2 :> y2=r27m2 :> yzx/r2—$2.sothe ‘ , surface area is S(;r) :27r(7"2 —a:2) +47ra: V13 —m2. 0 S m S 7“ : 27m"2 — 271'562 +47r($ V r2 — $2) Thus. S’(:z:) : 0 — 47m +47r [as %(r2 — m2)_1/2(—2w) + (r2 — 3:2)1/2-1] 2 7 2 _ 2 _ 2 2 7 ,2 :47r[,m,_$_+ T—zfi :4”. gun" at at +7“ 3L 7.2_$2 T2_x2 ...
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