Max value, closed interval

1. Dec 7, 2007

fitz_calc

1. The problem statement, all variables and given/known data

the max value of f(x)=x^3(40-x^2) on the closed interval 0<= x <= 40 occurs at x=?

3. The attempt at a solution

different problem, same story (notes are unclear, and the book only complicates things more). I know the answer is 24, but this will not help me on the final exam!

Can someone please provide a more clear explanation of what the book is asking? What would be the first step to determine the max value of x given the above constraints? thanks!

* i knwo i have to take the derivative of the equation, other than that im not so sure...

2. Dec 7, 2007

Mr.Brown

The first thing is to note that the function has to have a maximum and a minimum value on the intervall [0,40] since the intervall is compact.
Now there are two posibilities either the function takes it´s maximum in the intervall or on the boundary.
Now if the function takes the max in the intervall you just have to set the derivative equal zero find critical values in the intervall and you know .... :)
Now you have to calculate the boundary values in the end to see if the max is on the boundary perhaps because you can´t find these be derivatives :)

3. Dec 7, 2007

Defennder

I don't think the answer is 24. It should be x=sqrt(24).

To approach this problem, you only need to apply the 1st and 2nd derivative test for single variable function. Recall that the first derivative of a function gives you the gradient of the function at any defined x, right? Hence if you're looking for a maximum or minimum at say x=x1, you would agree that the f'(x1)=0 at x=x1, right? Note that if f'(x) is positive is means the function is increasing and negative implies that it is decreasing. So, we should expect that at the max point, it would neither be increasing nor decreasing, ie f'(x)=0. This is because the tangent line at that point would be horizontal ie. neither sloping downwards nor upwards because that is the highest point already.

To check that x1 is indeed the maximum, you apply the 2nd derivative test; differentiate the function twice and evaluate it at that point. Applying this test to x1, you either get a negative, 0 or positive answer. If it's negative, it means that the gradient of the graph at that point is is starting to decrease, which implies that point is a maximum, rather than minimum. If positive, it means the gradient is starting to increase, which implies that it is a minimum.

Hope this helps.

4. Dec 7, 2007

HallsofIvy

Staff Emeritus
If a function is continuous on a closed and bounded interval (compact, as Mr. Brown said), then it takes on both maximum and minimum values on that interval.

The maximum and minimum values must occur either in the interior of the interval (not an endpoint) where the derivative is equal to 0 or at an end point.

Since f(x)=x^3(40-x^2), f'(x)= 3x^3(40- x^2)+ x^3(-2x)= x^3(120- 3x^2- 2x)That will be equal to 0 when (x^3)(120- 2x- 3x^2)= 0. Find the roots of that equation, evaluate f at those values, at x= 0, and at x= 40. Whichever value is highest is the maximum value for f(x).