Max/Min Word Problems (Additional Review) Solutions. =, for 2 x 5 1 x 1 x ( 1) 1+ ( ) ( ) ( ) 2 ( ) x = 1 + (2) 3 1 (2) (5) (5) 4 2

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May Garrison
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1 . a) Given Ma/Min Wod Poblems (Additional Review) Solutions + f, fo 5 ( ) + f (i) f 0 no solution ( ) (ii) f is undefined when (not pat of domain) Check endpoints: + () f () () + (5) 6 f (5) (5) 4 (min. value) (ma. value) Thee is an absolute ma. pt. at b) Given g sin + cos, fo g cos sin (i) g 5, and an absolute min. pt. at (, ) cos sin cos cos sin. cos sin 0 (ii) g is defined fo all values of cos 0 o sin 0 in the domain, 6 Check values: g sin + cos sin + cos sin + cos g g (min. value) (ma. value) Thee is an absolute ma. pt. at (, ) and an absolute min. pt. at 6,.

2 . Let be the numbe of music playes sold. Let p be the pice pe playe. Given (, p ) ( 8000,50) and (, p ) ( 7900,5) 50 5 then m p 50 Fo the pice function, p p p Revenue function: R p R (i) R (ii) R is defined fo all values in the 50 domain (6500) $65 00 Find pice: p They should chage $65 pe playe in ode to maimize evenue.

3 . a) Given C , The aveage cost function is given by When 000 items ae poduced, c (000) (000) 000 $6.50/item c. The maginal cost is given by C When 000 items ae poduced, C (000) (000) $5.50/item b) Fo minimum aveage cost, find the deivative: c Solve c 0 : A poduction level of 000 items will minimize aveage cost. c) When 000 items ae poduced, c (000) (000) (000) $9.50/item The minimum aveage cost is $9.50/item when 000 items ae poduced.

4 4. a) Let be the width of the ectangle and let y be the length. shoeline Ma A y given + y 400 y 400 (sub into A y ) ( 400 ) A y 400 ( 0 00) A (i) A (ii) A is defined fo all values of in the domain 00 Check: A (0) 400(0) (0) 0 A (00) 400(00) (00) 0000 (ma.) A (00) 400(00) (00) 0 Find y: y 400 (00) 00 The maimum swimming aea is 0000 m when the dimensions ae 00 m 00 m b) Since the citical numbe fom pat a) is no longe pat of the domain ( 0 50) check new endpoints: A (0) 0 A (50) 400(50) (50) 5000 The new maimum aea is 5000 m when the dimensions ae 50 m 00 m.,

5 5. Let be the adius and h be the height of the can. Since the cost (C) of the mateials is based on the suface aea (A), we need to find the minimum value of h A + h given V h 000 cm 000 h (sub into A) 000 A A + ( > 0) 000 A (i) A 4 0 (ii) A is undefined when (but >0) Check using Second Deivative Test: 4000 A 4 + when 5.4, 4000 A (5.4) 4 + (5.4) > 0 a min eists at Find h: h (5.4) 0.84 To minimize the cost of the metal, the adius should be 5.4 cm and the height should be 0.8 cm

6 6. Let be the width of the ectangula potion of the window, and let y be the height of the ectangula potion. To admit the geatest amount of light, ma Aea Aea of ectangle + Aea of semi-cicle A y + (adius ) y 8 given Peimete y y 4 (sub into A) y A A (i) A (ii) A is defined fo all values of (4 + ) 8 in the domain 8 (4 + ). m Check: A (0) 8(0) (0) (0) 0 m A (.) 8(.) (.) (.) 4.48 m (ma) A (.56) 8(.56) (.56) (.56).79 m Detemine width (.).4 m To admit the geatest amount of light, the width of the window should be.4 m.

7 7. Let the length of the ectangle be and the width be y, such that (,y) is a point on the ellipse as shown. ma Aea y 4y y given + 4y 4 4 4y 4 4 y (sub into Aea equation) Aea A( y) 4y 4 4y ( 0 y ) A ( y) 4 4 4y + 4y 4 4y 8y 4( 4 4y ) ( 4 8y ) ( y ) 4 4y 4 4 4y 4 4y 4y 6 (i) ( y ) 4 4y 6 A ( y) 0 y 0 y y (ii) A ( y) is undefined when y Find dimensions: 4 4 y Check: A (0) 4(0) 4 4(0) 0 A (ma) A () 4() 4 4() 0 The lagest ectangle which can be inscibed in the given ellipse has an aea of 4 units when the dimensions of the ectangle is units units.

8 8. Let t be the time (in hous) since noon, when the boats ae closest togethe. Let d be the distance between them. 5t At :00 noon + t hous, the westbound boat has tavelled 5t km and 0 0t the nothbound boat has tavelled 0t km, so d it is 0 0t km fom the dock. Minimize d such that: d ( 5t ) + ( 0 0t ) 65t t + 400t t t ( 0 t ) dd Find the deivative: d 050t 800 dt dd 050t t 800 5t + 0 0t dt d dd 050t 800 (i) 0 dt 5t + 0 0t 050t t (ii) dd dt is defined fo all values of t in the domain Check: d (0) 05(0) 800(0) km d (0.9) 05(0.9) 800(0.9) km (min.) d () 05() 800() km The boats ae closest togethe at 0.9 hous afte noon, ie. at appoimately : pm. Note: minutes

9 9. Let (00 ) be the distance the cable should be un on land. Let y be the distance the cable should be un unde wate. Fom the diagam, y y Minimize cost such that C y ( 0 00) Find the deivative: C ( ) ( ) m P y 00 Q (i) C (ii) C is defined fo all values in domain Check: C (0) (0) + 80 (0) $56000 C (57.7) (57.7) + 80 (57.7) $ (min) C (00) (00) + 80 (00) $96.76 To minimize the cost of laying the cable, it should be installed fom Q to a point 4 m east, then 5.5 m unde wate to point P.

10 0. a) Define points E, F, G, and H as shown in diagam. Note: AEH CGF θ ( 0 θ ) Aea of ABCD AB BC ( AE + EB ) ( BF + FC ) B E F C cm θ 5 cm A H D G Fom AEH, Fom FCG, AE cos 5 θ FC sin 5 θ AE 5cos θ FC 5sin θ Fom EBF, BEF θ EB BF EB sin θ BF cos θ EB sin θ BF cos θ cos ( θ ) sin ( θ ) This step uses Complementay Identities (see pg. 9 of tet) Fom above, Aea of ABCD ( AE + EB ) ( BF + FC ) ( 5cos θ + sin θ)( cos θ + 5sin θ ) 5cos θ + 5sin θcos θ + 9sin θcos θ + 5sin θ 5 sin θ + cos θ + 4sin θcos θ 5() + 7(sin θcos θ ) A( θ ) sin θ (i) da 4cos 0 d θ θ cos θ 0 θ θ ad 4 da 4cos d θ θ (ii) da is defined fo all values in domain dθ

11 Check: A(0) sin[ (0) ] 5 cm A ( 4 ) sin ( 4 ) A sin 5 cm cm (ma) The maimum aea is cm when θ adians. 4 b) Fom pat a), in ectangle ABCD, length AE + EB width BF + FC When θ 4 5cos θ + sin θ cos θ + 5sin θ length 5cos + sin width cos + 5sin Fo maimum aea, the dimensions of ectangle ABCD should be 4 cm 4 cm.

CALCULUS FOR TECHNOLOGY (BETU 1023)

CALCULUS FOR TECHNOLOGY (BETU 103) WEEK 7 APPLICATIONS OF DIFFERENTIATION 1 KHAIRUM BIN HAMZAH, IRIANTO, 3 ABDUL LATIFF BIN MD AHOOD, 4 MOHD FARIDUDDIN BIN MUKHTAR 1 khaium@utem.edu.my, iianto@utem.edu.my,