Question

The ordering and transportation cost C for components used in a manufacturing process is approximated by C(x) = 20 (1/x + x/ (x+3)) where C is measured in thousands of dollars and x is the order size in hundreds. (a) Verify that C(3) = C(6). (b) According to Rolle’s Theorem, the rate of change of the cost must be 0 for some order size in the interval (3, 6). Find this order size. Round your answer to three decimal places.

Answer 1

C(x)=20[(x+3)+x²)]/(x(x+3))

we need to know if c(3)=c(6)
Nothing more than a direct substitution

When you simplified c(x) it boils down to

C(x)=20[(x+3)+x²)/((x)(x+3)]

Where x=3

C(3)=20[(6+9)/(3(6)=20(3(2+3)/3(6)
=20(5/6)=10(5/3)

Where x=6

C(6)=20(9+36)/6(9)=20[(9(1+4)/9(6)
=20(5/6)=10(5/3)

Which satisfy the claim that c(3)=c(6)

From
C(x)=20[(x+3)+x²)/(x²+3x)

Using quotients rule

V=(x²+3x) & u=(x+3)+x²

V'=2x+3. u'=2x+1

C'x=20(vu'-uv')/v²

C'(x)=20[(x²+3x)(2x+1)-(((x²+x+3)(2x²+3))]/(x²+3x)

The value for x for which c'(x)=0

20[(x²+3x)(2x+1)-(((x²+x+3)(2x²+3))]/(x²+3x)=0


[(x²+3x)(2x+1)-(((x²+x+3)(2x²+3))]=0

[2x³+6x²+x²+3x-(2x³+2x²+6x+3x²+3x+9)=0


2x³+7x²+3x-2x³-5x²-9x-9=0

2x²-6x-9=0

a=2 b=-6 c=-9

x=(6±√(36-4(2)(-9)))/4

x=6±√(36+72)))/4

x=[6±√9(4+8))]/4

X=[6±3√(4+8)]/4

x=[6±3√4(1+2)]/4

x=[6±6√3]/4

x=6(1±√3)/2

X=3(1±√3)/2

It must be a positive size so between

3(1-√3)/2 and 3(1+√3)/2

Which would it be?

Answer 2

For the entries

C'(x)=20[(x²+3x)(2x+1)-(((x²+x+3)(2x+3))]/(x²+3x)²

When you expand it is still the same I didn't really check it this work b4 post it

Was doing alot of copying and pasting so no worries.....

Answer 3

Correction again
Typo
x=[6±6√3]/4≠6(1±√3)/2

But
x=6(1±√3)/4=3(1±√3)/2