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Chapter 4 96 - 358[3 CHAPTER4 APPLICATIONS OF...

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Unformatted text preview: 358 [3 CHAPTER4 APPLICATIONS OF DIFFERENTIATION 9. 10. 11. 12. 13. 14. my = 1.5 x 106. so y = 1.5 X 106/37. Minimize the amount of fencing. which is 3s + 2y = 3x + 2(1.5 X 10%) = 37; + 3 x 106/2: : Fm). F’(m) = 3 — 3 x 106/37:2 2 3(102 — 106) /ac2. The critical number is x :103 and F'(w) < 0for0 < x < 103 and F’(:c) > 0 ifm > 103. so the absolute minimum occurs when x = 103 and y = 1.5 X 103. The field should be 1000 feet by 1500 feet with the middle fence parallel to the short side of the field. Let b be the length of the base of the box and h the height. The volume is 32.000 : bzh => h = 32.000/b2. The surface area of the open box is S : b2 + 4hb = b2 + 4(32.000/b2)b : b2 + 4(32.000)/b. So S’(b) : 2b — 4(32,000)/b2 : 2(b3 — 64000) /b2 : 0 4:» b : W : 40. This gives an absolute minimum since S’(b) < 0 ifO < b < 40 and S'(b) > 0 ifb > 40. The box should be 40 X 40 X 20. Let b be the length of the base of the box and h the height. The surface area is 1200 = 02 + 4hb => h : (1200 — b2)/(4b). The volume is V 2 02h : b2(1200 , b2)/4b : 300b , 123/4 :> V’(b) = 300 — 3b? V’(b) 0 300 g1)? ? b2 400 0 Wm 20. Since V’(b) > 0 for 0 < b < 20 and V’(b) < 0 for b > 20. there is an absolute maximum when b : 20 by the First Derivative Test for Absolute Extreme Values (see page 334). If I) = 20. then It : (1200 — 202)/(4 - 20) = 10. so the largest possible volume is b2h : (20)2(10) : 4000 cm3. - V : lwh => 10 : (2w)(w)h : 2w2h. so It : 5/1112. The COSt is 10(21122) + 6[2(2wh) + 2(hw)] : 20w2 + 36wh. so C(w) : 20w2 + 36w(5/w2) : 20w2 + 180/11). 2w C'(w) : 40w 7 180/11)2 2 40 (103 7 gym? :> w : f/gis the critical number. There is an absolute minimum for C when w : i/g since C'(w) < 0 for 0 < w < {/E and C(20) >0forw> {E OWE) :20(§/§)2+ £0? m$163.54. - 10 : (2w)(w)h : 2102b. so h : 5/102. The cost is C(w) = 10(2w2) + 6[2(2wh) + 2hw] + 6(2w2) w : 32w2 + 36wh : 32w2 + 180/212 2w C'(w) : 64w — 180/102 : 4 (167.03 7 45)/w2 2 w : ,3/i11—g is the critical number. C'(w) < 0 for 0 < w < 3 $5; and C’(w) > 0 form > 3 %. The minimum cost is C( 3 fi) : 32(2.8125)2/3 +180/x/2.8125 % $191.28. (a) Let the rectangle have sides m and y and area A. so A : any or y : A/m. The problem is to minimize the perimeter : 2:5 + 23; : 23: + 2A/m : P(a:). Now P'(m) : 2 — 214/232 : 2(cc2 — A) /a:2. So the critical number is m : x/Z. Since P'(w) < 0 for 0 < at < fl and P'(;c) > 0 for at > x/Z there is an absolute minimum at m : x/Z The sides of the rectangle are x/Z and A/x/Z : x/A7 so the rectangle is a square. (b) Let p be the perimeter and a: and y the lengths of the sides. so 1) : 2x + 2y :> 2y : p — 2m => y : ép— 3:. The area is 14(30): $619 — $) : %p$ — $2. Now A'(1:) : 0 => %p — 2a: : 0 => ...
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