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.

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?

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.....

Correction again

Typo
x=[6±6√3]/4≠6(1±√3)/2

But
x=6(1±√3)/4=3(1±√3)/2

To verify that C(3) = C(6), we need to substitute 3 and 6 into the equation C(x) = 20 (1/x + x/ (x+3)) and check if the results are equal.

(a) Calculate C(3):
C(3) = 20 (1/3 + 3/6)
= 20 (1/3 + 1/2)
= 20 (2/6 + 3/6)
= 20 (5/6)
= 16.667

(b) Calculate C(6):
C(6) = 20 (1/6 + 6/9)
= 20 (1/6 + 2/3)
= 20 (6/36 + 24/36)
= 20 (30/36)
= 16.667

Since C(3) = C(6) = 16.667, the verification is correct.

Now, let's find the order size in the interval (3, 6) where the rate of change of the cost is 0, according to Rolle's Theorem. To find this order size, we need to differentiate the cost function C(x) and find the root of the derivative in the interval (3, 6).

The derivative of C(x) is found by differentiating each term separately using the power rule:

C'(x) = 20 (-1/x^2 + (x + 3 - x)/(x + 3)^2)
= 20 (-1/x^2 + 3/(x + 3)^2)

Now, we need to find the value of x where C'(x) = 0.

Setting C'(x) = 0, we have:

0 = -1/x^2 + 3/(x + 3)^2

Multiplying through by x^2 and (x + 3)^2 to eliminate the denominators, we get:

0 = (x + 3)^2 - 3x^2

Expanding and rearranging the equation, we have:

0 = x^2 + 6x + 9 - 3x^2

Simplifying further, we have:

0 = -2x^2 + 6x + 9

Next, we'll solve this quadratic equation by factoring or using the quadratic formula.

Since this quadratic equation can't be factored easily, we'll use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -2, b = 6, and c = 9. Substituting these values into the quadratic formula, we get:

x = (-6 ± √(6^2 - 4(-2)(9))) / (2(-2))
x = (-6 ± √(36 + 72)) / (-4)
x = (-6 ± √108) / (-4)
x = (-6 ± 3√12) / (-4)

The two possible solutions are:

x = (-6 + 3√12) / (-4)
x ≈ 0.518 (rounded to three decimal places)

x = (-6 - 3√12) / (-4)
x ≈ -4.018 (rounded to three decimal places)

Since the order size cannot be negative, the only valid solution in the interval (3, 6) is approximately x = 0.518.