# Thread: Need help finding maximum of y = -4x^3 - 27x^2 + 5x +45 on -6<=x<=5

1. ## Need help finding maximum of y = -4x^3 - 27x^2 + 5x +45 on -6<=x<=5

Hi, I need help in pointing me to the right direction on how to solve this problem.

The question: Find the maximum of y = -4x3 - 27x2 + 5x +45 in the range -6 <= x <= 5.

I believe this is a cubic equation and I am having trouble where to start on solving this problem.
Is there a way to solve this problem without using derivative, if so, could you tell me how.
It would be best if I could solve this without derivative.

Thank you!

2. You are in a calculus class. One may wish to try the 1st derivative and then check the end points. This should have been covered in class and in the text. No?

3. Originally Posted by cheapchickens
Is there a way to solve this problem without using derivative, if so, could you tell me how.
You could guess. Then determine whether your guess is too big or small, and, repeat …

It would be best if I could solve this without derivative.
Why? Using the Power Rule is easy, to find the first derivative. After that, the Quadratic Formula gives two candidates to check.

4. Originally Posted by cheapchickens
Hi, I need help in pointing me to the right direction on how to solve this problem.

The question: Find the maximum of y = -4x3 - 27x2 + 5x +45 in the range -6 <= x <= 5.

I believe this is a cubic equation and I am having trouble where to start on solving this problem.
Is there a way to solve this problem without using derivative, if so, could you tell me how.
It would be best if I could solve this without derivative.

Thank you!
Not easily nor without some work, much more than the simple 'take the derivative'. If I really had to do it without derivative, I would, as mmm4444bot said, guess and correct. Before starting that thought, I might go through some reasoning such as: First look at the equation. y goes to infinity as x goes to minus infinity and minus infinity as x goes to infinity. This means it has either 1 [0] or 3 [2] zeros [extrema]. Now try to get an image of the graph in your mind on the interval [-6, 5]: It is negative at the two end points of the interval (you need to do the computations) and positive at x=0. So there are two zeros in the interval indicating a single maximum. [Note that since y is negative at -6 and goes to infinity as x goes to infinity, y has a minimum some where in the interval also.] Thus we have accounted for all three zero of y and the maximum extrema [the one between x=-6 and x=5].

Now, off to work [some more computations]:
(1) dx=1, i.e. compute y at -6, -5, -4, ...
At x=-1, y=17 [notation (-1,17)], (0,45), (1,19). Going up at x=-1, down at x=0, so max is between -1 and 1.
(2) dx=.5 starting at -0.5 [you already have the value at -1]
Going up at x=-0.5, down at x=0.5, so max is between -0.5 and 0.5.
(3) ....

As I said, a lot more work.

5. 1) Why? Thus my initial response.
2) We have graphing facilities galore - calculators, CAM, online apps.
3) This is why we invented the calculus. Shall we get Dino the Dinosaur to teach your calculus class?

Your ONLY recourse, without the calculus, is some sort of visualization. Exactly how you choose to produce such a visualization is of no consequence.

6. Originally Posted by cheapchickens
Hi, I need help in pointing me to the right direction on how to solve this problem.

The question: Find the maximum of y = -4x3 - 27x2 + 5x +45 in the range -6 <= x <= 5.

I believe this is a cubic equation and I am having trouble where to start on solving this problem.
Is there a way to solve this problem without using derivative, if so, could you tell me how.
It would be best if I could solve this without derivative.

Thank you!
As the other answers have said, this APPEARS to be a silly question unless you are studying numerical methods. Calculus easily solves it.

If you are studying numerical methods, what have you tried?

7. ## Don't give up.

Ishuda showed that -0.5 < x < 0.5.
He said, "a lot more work."

Actually, you are almost there (if all you need is a few significant digits).
P(x)-45 = -4xxx -27xx + 5x has maximum at same x value.

Over
-0.5 < x < 0.5. the term -4xxx is insignificant.
P(x)-45 ~ -27xx + 5x
P(x)-45 ~ x(5-27x)

Zeros of quadratic at 0 and 5/27
Max of
quadratic is halfway between the zeros.
Max at x ~ 5/54 = 0.09

8. Even the common Numerical Methods tend to exploit principles from the Calculus (continuity, mean value, slope, and on and on...)