Question

Continuing from the previous related rates problem..

Answer 1

The manager of a company determines that when q hundred units of a particular commodity are produced, the total cost of production is \(C\) thousand dollars, where \(C^2−3q^3=4,275.\) When 1500 units are being produced, the level of production is increasing at the rate of 20 units per week. What is the total cost at this time and at what rate is it changing?

Answer 2

We solve for the cost..
\[\LARGE C^2-3q^3=4275\]
\[\LARGE C=\sqrt{3q^3+4275}\]

Answer 3

yes plugin q = 15,
that gives the total cost when 1500 units are produced

Answer 4

note that q is in hundreds

Answer 5

And then we derive it for the change?

Answer 6

it is better to derive the original equation for change

Answer 7

\(\large C^2−3q^3=4,275 \)

differentiate both sides with respect to time

Answer 8

\(\large \frac{d}{dt}( C^2−3q^3=4,275) \)

Answer 9

\(\large 2C\frac{dC}{dt}−9q^2\frac{dq}{dt}=0\)

Answer 10

plugin known values and u can find \(\frac{dC}{dt}\)

Answer 11

just be careful when plugging in q and dq/dt values
u need to change units to hundreds and plugin.. cuz q is in hundreds

same applies to C also, C is in thousands...

Answer 12

Wouldn't it also work to plug in \(\large \frac{dq}{dt}=.2\) when \(q=15\)

Answer 13

exactly !

Answer 14

thats what i meant, we need to plugin them and not 20 and 1500

Answer 15

The numbers are big enough as they are huh? xD
Okay, so I have..
\[\LARGE 2(.12)\frac{dC}{dt}-9(15)^2(0.2)=0\]
Just solve this for \(\large \frac{dC}{dt}\)?

Answer 16

hey C = 120, not 0.12

Answer 17

you have solved the equation in thousands oly,
so u wud get the value of C in thousands oly

Answer 18

C = 120000 units = 120 K units
so we need to plugin C = 120

Answer 19

Whoops, guess I didn't need to convert on that.. thanks for the catch :)

Answer 20

np :) yes just solving for dC/dt will do..

Answer 21

So..
\[\LARGE \frac{dC}{dt}=1.6875\].-.

Answer 22

wat about units ?

Answer 23

1.6875 thousand dollars per week ?