Math Solving Problem Derivation

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MATH SOLVING PROBLEM “THE APPLICATION OF DERIVATION” 3rd Group : Anita Rosalina Nadia Comeneci Ratih Surya Pratiwi

XI IIA 2 INTERNATIONAL CLASS SMAN 8 PEKANBARU 2008/2009 1

Question 4. Find the least amount of material needed to build an open right cylindrical vessel with capacity of 400π cm3.

TRANSLATE PLEASE!!! 2

• Temukan jumlah terkecil material yang dibutuhkan untuk membuat sebuah bejana silinder terbuka dengan kapasitas 400 π cm3

3

ANSWER V = πr t 2

400π = πr t 2

400 t= 2 r

SA = 2πrt +πr 2  400  SA = 2πr  2  +πr 2  r  SA =800πr −1 +πr 2 dSA = −800πr −2 +2πr dr 0 = −800πr −2 +2πr 800πr −2 = 2πr r = 400 −2 r r 3 = 400 r = 3 400

4

SA = 2πrt + πr 2 400 t= 2 r 400 t= 3 400 400 t= 2

(

400 3 t = 400

( SA = 2π ( SA = 3π ( SA = 2π

)

2

3 3

3

)( 400 ) + π ( 400 ) 400 ) + π ( 400 ) 400 ) 400

3

2

3

3

2

2

2

SA = 162,87πcm

2

1 3

t = 3 400

So, the least amount of material needed is 162,87 π cm2 5

QUESTION 5. A rectangular box has a square base of side x cm. If the sum of one side of the square and the height is 15 cm, express the volume of the box in terms of x. Use this expression to determine the maximum volume of the box.

TRANSLATE PLEASE!!! 6

• Sebuah balok mempunyai alas persegi dengan sisi x cm. Jika jumlah salah satu sisi persegi dan tinggi adalah 15 cm, nyatakan volum balok dalam x. Gunakan persamaan ini untuk menentukan volum maksimum dari balok.

7

ANSWER x + t = 15cm t = 15 − x

V = x 2t

V = x 2 (15 − x ) V =15 x 2 − x 2

t = 15 − x t = 15 − 10 t = 5cm V =x t 2

V ' = 30 x − 3 x 2 0 = 30 x − 3 x 2 3 x 2 = 30 x x =10cm

V = (10 ) ( 5) 2

Vmaks = 500cm 3

8

QUESTION 6. A sector of a circle with radius r cm contains an angle of θ radian between bounding radii. Given that the perimeter of the sector is 7 cm, express θ in terms of r and show that the area of the sector is 1 r (7 −2r ) cm 2

2

TRANSLATE PLEASE!!!

Hence or otherwise, find the maximum area of this sector as r varies. 9

• Sebuah juring dari lingkaran dengan jari-jari r cm memiliki sudut θ radian diantara jari-jari yang terhubung. Garis kelilingnya ialah 7 cm, nyatakan θ dalam r dan tunjukkan bahwa luas 1 r (7 − 2r ) cm juring ini adalah 2 Sebaliknya, temukan luas maksimum dari juring ini. 2

10

ANSWER r x r

2r +x =7cm 2r +

α 360°

2πr =7

α 2r + 2πr =7 2π 2r +αr =7 αr =7 −2r α=

7 −2r r

α SecA = πr 2 2π αr 2 SecA = 2 2 7 − 2 r r   SecA =    r  2 1 SecA = r ( 7 − 2r ) 2 SHOWED!!!! 11

7 SecA = r − r 2 2 7 SecA' = − 2r 2 7 0 = − 2r 2 7 2r = 2 4r = 7 7 r= 4

7 SecAmaks = r − r 2 2  7  7   7  SecAmaks =    −    2  4   4  49 49 SecAmaks = − 8 16 98 − 49 SecAmaks = 16 49 SecAmaks = 16 SecAmaks = 3,0625cm 2

2

12

A M RI TE

SI KA

H

N A TH

U O KY

13

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