Math question help? Please I have an exam tommorrow!

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DIFFERENTIATION QUESTION. PLEASE HELP. I HAVE NO CLUE ON HOW TO SOLVE THESE EXCEPT THE FIRST ONE a) i) BY SPEED = DISTANCE/TIME. THAT'S ALL I KNOW.

A ship has a 200km journey to make at a constant speed. At x km/h the cost, in $, of the journey will be:

(x^2 + 4000/x ) PER HOUR.

a)Find an expression for

i)the time taken for the journey
ii) the total cost of the trip

b). Find the speed that minimses the cost of the journey and calculate this minimum cost.

THE ANSWWERS ARE

a). i) t= 200/x
ii) 200x + 800000/x^2
b). 20 km/h, cost is $6000

Well the total cost of the journey is simply the cost expression multiplied by the total time, which is

[x^2 + 4000/x] * [200/x]

or 200x + 800,000/x^2

Then you want to differentiate that function in order to optimize it. That expression is

200 - 1,600,000/x^3

In this case, you can simply set that expression equal to zero in order to find the optimal value

So

200 - 1,600,000/x^3 = 0

200 = 1,600,000/x^3

200x^3 = 1,600,000

x^3 = 8,000

Then the cube root of both sides is 20.

Let me know if you don't understand anything here

One little point, chever...

Having done the differentiation, and solved for a turning point of the function, you must verify that against any boundary values, etc.

In this case, the "obvious" things to check are x -> 0 and x -> inf, which both yield cost->inf, so that's OK.

The solution, x = 20, gives a cost of 6,000. which seems fine.

However, being pedantic, a speed of -inf gives a cost of -inf. I.e. don't go toward your destination, but run away from it, as fast as you can, and you will earn money.

The trouble with models, is that you always have to be sure you haven't overlooked something.

Of course. That's why I said "In this case, you can simply set that expression equal to zero in order to find the optimal value"

Most optimization questions in an introductory calculus class will not throw curve balls at you like that, and this one clearly does not have such a curve ball.

SO HOW DO YOU GET 6000?

Plug 20 into the total cost function.

I hate to sound brusque, but you will have a lot of difficulty with introductory calculus, to say nothing of the limits / integration techniques, vector calculus and diff eq's you might take later, if you don't master these techniques.

Not to mention integration is a little trickier then differentiation.

But for the problem stated by the original poster the given method works. The curve balls come in Calculus II & III (Integration & Multi-variable).

I managed to get a C in Calc II.

Looks like I'm going to get an A in Calc III.

Go figure.

Yeah I got roughly the equivalent of a C in Calculus II & then obtain an AB in Calculus III (working my butt off does help though & I managed to do this without the vaunted TI-89)