# Math Help - how to find maximum and minimum points for this equation?

1. ## how to find maximum and minimum points for this equation?

(pi)x^2 + 4(pi)^2.x^2 - 28(pi).x.a + 49a

(a=tan72)

the first derivative according to me is... 2(pi)x + 8(pi)^2.x.a-28(pi)a

ive been trying to make x the subject of this equation, when i do get the x, what do i do with it???

2. Originally Posted by saha01
thanks for the reply, i got x=1.5989, and for the second derivative i got 2pi + 8(pi)^2a, which is equal to -14.44, and i dont really undertand what this value means...is it the maximum? if it is, how do i find the minimum value?
You don't need to use the second derivative here. I was talking about zeroes, which are different.

Given $f(x)=ax^2+bx+c$, there are two zeroes such that $f(x_1) \text{ and}f(x_2)=0$. That's what you want to find for f'(x)/ x_1 and x_2 such that those numbers equal 0. I would suggest the quadratic formula for the easiest way.

If f'(x)=0, that means the slope is 0. This means the graph is flat at this point and very likely is changing directions. With me so far?

3. (pi)x^2 + 4(pi)^2.(x^2).a - 28(pi).x.a + 49a

this is the equation i need to differentiate

4. Originally Posted by saha01
hmmm, this is the original equation, not the derivative...
Oh. I just cannot read tonight, sorry again.

Then $f'(x)=(2\pi+8a{\pi}^2)x-(28a\pi)$

Set that equal to zero then solve for the critical points. Do you know what to do from there or want more guidance?

5. Originally Posted by Jameson
Oh. I just cannot read tonight, sorry again.

Then $f'(x)=(2\pi+8a{\pi}^2)x-(28a\pi)$

Set that equal to zero then solve for the critical points. Do you know what to do from there or want more guidance?
thanks, now i got x=1.598968004, what am i to do with his value? do i put it back into the original equation? or do i take a second derivative?

6. Originally Posted by saha01
thanks, now i got x=1.598968004, what am i to do with his value? do i put it back into the original equation? or do i take a second derivative?
None of those. Draw a number line and mark the point you got on it. Now take any number before, meaning less than, that and see if it is positive or negative. Do the same for a number after, meaning greater than, that number. If you see a change in sign of f'(x) at that number, it is a critical point.

If it goes from - to +, you have a local minimum. If it goes from + to - you have a local maximum.

7. ok.... so since my x value was 1.5989....

i chose a number greater than x, which was 2, and plugged it into the derivative, and got a negative number, which means that this point is a maximum.

but when i chose a number less than my x value, i get a positive number everytime, what happens if there is no sign change?

i'm sorry if im causing you any trouble btw. and thanx a lot, ur help really means a lot

8. Originally Posted by saha01
ok.... so since my x value was 1.5989....

i chose a number greater than x, which was 2, and plugged it into the derivative, and got a negative number, which means that this point is a maximum.

but when i chose a number less than my x value, i get a positive number everytime, what happens if there is no sign change?

i'm sorry if im causing you any trouble btw. and thanx a lot, ur help really means a lot
Are you in degrees or radians for this problem? I tried it in degrees and got the answer I think it should be, which is that you have a local minimum. Try it again with your calculator in degrees.

If you did get a change from positive to negative though you are right it's a maximum. If there is no change it is called an inflection point, but don't worry about that yet.

The reason I think it should be a minimum is that you started with a parabola. Since the number in front of x^2 was positive, it should open upwards meaning it only has a minimum value but no maximum.

9. when i plugged in 2 again in degree mode, i got +227.8474...

and with 1.2 (smaller no. than x) i got 28.417...

10. Ok I just need to do this and end this time consuming back and forth. I'm sorry I haven't been that helpful. . I just noticed a problem. In your original post you wrote $4\pi ^2 x^2$ but later on you said it was $4\pi ^2 ax^2$. This will definitely make a difference.

Assuming you mean the second one, this is how to solve for x.

$
f'(x)=(2\pi+8a{\pi}^2)x-(28a\pi)=0
$

$(2\pi+8a{\pi}^2)x=(28a\pi)$

$x=\frac{28a\pi}{2\pi+8a{\pi}^2}$

Clear up this last bit of info and I'll finish the problem!

11. Originally Posted by Jameson
Ok I just need to do this and end this time consuming back and forth. I'm sorry I haven't been that helpful. . I just noticed a problem. In your original post you wrote $4\pi ^2 x^2$ but later on you said it was $4\pi ^2 ax^2$. This will definitely make a difference.

Assuming you mean the second one, this is how to solve for x.

$
f'(x)=(2\pi+8a{\pi}^2)x-(28a\pi)=0
$

$(2\pi+8a{\pi}^2)x=(28a\pi)$

$x=\frac{28a\pi}{2\pi+8a{\pi}^2}$

Clear up this last bit of info and I'll finish the problem!

yes thats what i got for x

12. Originally Posted by saha01
yes thats what i got for x
Ok then I get the same value you did, while in radians. I also noticed that tan(72) in radians is negative, so the coefficient of x^2 is negative, meaning the parabola flips down, not up.

You are correct that the point you found is a local maximum. To get the actual point remember to plug x into the original equation, not the derivative. Sorry it took so long but you finally have the correct solution.

13. since they've asked for a minimum as well, shall i write doewn that a minimum does not exist? and to get the actual point, do i plug in 1.5989....into the original equation?

14. Originally Posted by saha01
since they've asked for a minimum as well, shall i write doewn that a minimum does not exist? and to get the actual point, do i plug in 1.5989....into the original equation?
There is no minimum, correct. And yes as well to the second part. Plug in 1.5989 into f(x), not f'(x).

15. ok...i got 5.5963....which seems correct. thanks a million for your help!