Cambridge Additional Mathematics

0

x

• 
  
  
  
  • 20
  
  
  
  

A_'(x)

xcm

ycm

xcm

open

DEMO

+ -

5

0 12.5

x

d

d__Vv~

394 Applications of differential calculus (Chapter 14)

Step 3: dV

dx

=12x^2 ¡ 260 x+ 1000

= 4(3x^2 ¡ 65 x+ 250)

= 4(3x¡50)(x¡5)

)

dV

dx

=0 when x=^503 =16^23 or x=5

Step 4:

dV

dx

has sign diagram:

Step 5: There is a local maximum whenx=5. This is the global maximum for the given domain.

Step 6: The maximum volume is obtained when x=5, which is when 5 cm squares are cut

from the corners.

Example 19 Self Tutor

A 4 litre container must have a square base, vertical sides,

and an open top. Find the most economical shape which

minimises the surface area of material needed.

Step 1: Let the base lengths bexcm and the depth beycm.

The volume V=length£width£depth

) V=x^2 y

) 4000 =x^2 y .... (1) f 1 litre ́ 1000 cm^3 g

Step 2: The total surface area A=area of base+4(area of one side)

=x^2 +4xy

=x^2 +4x

³

4000

x^2

́

fusing (1)g

) A(x)=x^2 + 16 000x¡^1 where x> 0

Step 3: A^0 (x)=2x¡16 000x¡^2

) A^0 (x)=0 when 2 x=

16 000

x^2

) 2 x^3 = 16 000

) x=

p 3

8000 = 20

Step 4: A^0 (x) has sign diagram:

If x=10,

A^0 (10) = 20¡16 000 100

=20¡ 160

=¡ 140

If x=30,

A^0 (30) = 60¡16 000 900

¼ 60 ¡ 17 : 8

¼ 42 : 2

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100 100 4037 Cambridge
Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_14\394CamAdd_14.cdr Wednesday, 9 April 2014 1:07:47 PM BRIAN