Find maximum of the function N(t) without Calculus

AI Thread Summary
To find the maximum of the function N(t) without calculus, the discussion emphasizes using algebraic techniques, particularly completing the square for the quadratic component t^2 - 48t. The symmetry of the function suggests that the maximum occurs at t = 24, yielding N(24) = 14648.13868. It is noted that if the function is quadratic, the vertex represents either a maximum or minimum, determined by the coefficients. The conversation also highlights the importance of accurately representing the function to avoid misinterpretation of its behavior. Overall, algebraic manipulation and understanding of parabolic properties provide a viable solution without calculus.
Mathman2013
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Homework Statement
Find maximum of N(t) without Calculus
Relevant Equations
N(t) = 1456*0.996^(t^2 - 48*t)
How would you go about finding maximum value for this function without Calculus? You can draw in it a CAS tool like Geogebra, NSpire or Maple. And use the maximise ability. But is possible to do it by hand? Pre-Calculus?
 
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PF guidelines ask you to post an attempt at solution before we can help.

When is something like this at a maximum ?

##\ ##
 
BvU said:
PF guidelines ask you to post an attempt at solution before we can help.

When is something like this at a maximum ?

##\ ##
I have tried to say I take the internal function y=t^2-48*t and solve y =0, then t^2-48*t = 0, then I get t = 0 or t = 48, and it looks to be the function is symetrical, so if I divide by 48 by 2, I get t = 24, and N(24) = 14648.13868 which looks to be a maximum. But is there a better way?
 

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caz said:
This is similar to maximizing 0.996^x. What does this say about x?
if x increases then N decreases?
 
caz said:
If you want to maximize N, what do you want x to do?
x to increase to. But why do I get the maximum point if solve the internal function as eqn: t^2-48*t=0 and divide one of the roots by two? Is it only do to symetry?
 
Mathman2013 said:
I have tried to say I take the internal function y=t^2-48*t and solve y =0, then t^2-48*t = 0, then I get t = 0 or t = 48, and it looks to be the function is symetrical, so if I divide by 48 by 2, I get t = 24, and N(24) = 14648.13868 which looks to be a maximum. But is there a better way?

I love the conditioned reaction to solve y = 0 and the ten digit accuracy of N(24), but unfortunately both are irrelevant here. The direct questions by @caz are more to the point.

If you want to maximize ##a^x## for a number ##a## between 0 and 1, what do you want for ##x## ? that it is zero, at a minimum, or at a maximum ?PS: @caz : in PF, we don't give the answer on a silver plate, but try to let the poster find his/her way towards the solution by him/herself using gentle nudges and inviting questions ...

##\ ##
 
The form ##y=ax^2+bx+c ## is a parabola whose axis of symmetry can be found by completing the square. That's the algebraic way of coming up with the number ## t=24 ## that you did.
 
Mathman2013 said:
x to increase to. But why do I get the maximum point if solve the internal function as eqn: t^2-48*t=0 and divide one of the roots by two? Is it only do to symetry?
Yes, if ##y=f(x)##, if it’s quadratic (something like ##f(x)=A x^2 + Bx + C## where ##A, B, C## are constants), then the point halfway between two zeros of ##f## will be either a maximum or a minimum.
 
Maybe worth noting that in the quadratic formula, the axis of symmetry (which gives the x co-ordinate of the min/max) is given by the part of the formula not under the square root sign.
 
  • #10
I think the OP left off a minus sign for ## N(t) ##. Otherwise the function that he has will have a minimum and not a maximum. (His graph in post 3 is also incorrect otherwise).
 
  • #11
Charles Link said:
I think the OP left off a minus sign for ## N(t) ##. Otherwise the function that he has will have a minimum and not a maximum. (His graph in post 3 is also incorrect otherwise).

No, the function is presented $$N(t)=1456 \cdot 0.996^{t^2-48 \cdot t}$$, where $$t \geq 0$$
 
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  • #12
My mistake. I didn't see the ^ in the OP. In any case ## t^2-48t=(t-24)^2-576 ##.
To do the arithmetic of ## .996^{-576} ## is the next problem. One possibility would be to use logarithms.
 
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  • #13
For this equation, you indeed don't need calculus to know t that will maximize N(t). If you add a simple algebraic step, the answer will be shown more clearly.
 
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  • #14
You divide by 2 because this is the way to complete the square and this way find the vertex of the parabola which is a max or a min.
 
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  • #15
Charles Link said:
One possibility would be to use logarithms.
How about the binomial expansion?
 
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