Higher maths further calculus questionnn . . .?

Question

In a small factory the production cost, C, in pounds, of assembling x components in a month is givem by

C(x) = (40x + 2400 / x) ^2 , x = 0

Calculate the minimum cost of production in any month and the corresponding number of components that are required to be built.



Thannxxxxx everyone x x x x

In a small factory the production cost, C, in pounds, of assembling x components in a month is givem by

C(x) = (40x + 2400 / x) ^2 , x = 0

Calculate the minimum cost of production in any month and the corresponding number of components that are required to be built.


Ive already tried everything that qspeechc has said and i cant get the answer thats y am using answers. i dont think youd need to expand the expression either coz its a chain rule. someone helllp! lol x


Thannxxxxx everyone x x x x

Answer 1

Expand the expression, put the derivative equal to zero. this will give you the number of components required to build, only positive whole numbers. Plug those values back into the orginal equation to get the cost. If the x value is not whole, take the numbers above and below and check both for a minimum cost.

Answer 2

C(x) = (40x + 2400/x).(40x + 2400/x)
C(x) = 1600.x² + 192000 + (2400)².x^(-2)
C `(x) = 3200.x + (-2).(2400)².x^(-3)
C`(x) = 0 for minimum value.
3200x^(4) - 2 x 2400² = 0
x^(4) = 2 x 2400² / 3200
x^(4) = 3600
x = 8 is number of components that have to be built (to nearest whole number)
Minimum cost is when x = 8
C(8) = (320 + 300)²
C(8) = 384400
Minimum cost = £384,440

Hope this is aw right hen---good luck with H Maths!

Answer 3

Find the derivative of C(x). But, before you can do so, expand you equation 1st.
C(x) = 1600x^2 + 96000 + 5760000/(x^2)
C'(x) = 3200x - 11520000/(x^3)
C'(x)=0
3200x - 11520000/(x^3) = 0
3200x^4 = 11520000
x^4 = 3600
x = 7.75 components required
Minimum cost = 384000.09

Answer 4

C(x) = (40x + 2400 / x) ^2

For minimum cost, C'(x) = 0, so:
C(x) = 1600x^2 + 192000 + 5760000x^(-2)
C'(x) = 3200x + 0 - 11520000x^(-3)

C'(x) = 0
3200x - 11520000x^(-3) = 0
3200x = 11520000x^(-3)
3200x^4 = 11520000
x^4 = 3600
x = 7.75 (to 3 sig fig.)

Check:
C''(x) = 320 + 34560000x^(-4) > 0 => minimum cost.

Minimum cost,
C(7.75)
= (40(7.75) + (2400/7.75)) ^2
= 384,000 pounds

Answer 5

y = one million/x³ - cos2x y = x-³ - cos2x dy/dx = -3x^-4 - (-Sinx * 2) dy/dx = -3x^-4 + 2Sin2x interior the 2d area, the x comes difficulty to 2 factors, the Cos x & the 2x, so it is going by using a double technique of being differaterated.