Find maximum of the function N(t) without Calculus

[Mathman2013]

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?

 

[BvU]

PF guidelines ask you to post an attempt at solution before we can help.

When is something like this at a maximum ?

$\$

 

[Mathman2013]

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?

 

[Mathman2013]

This is similar to maximizing 0.996^x. What does this say about x?

if x increases then N decreases?

 

[Mathman2013]

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?

 

[BvU]

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 ...

$\$

 

[Charles Link]

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.

 

[stevendaryl]

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.

 

[neilparker62]

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.

 

[Charles Link]

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).

 

[Mathman2013]

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 $$t \geq 0$$

 

[Charles Link]

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.

 

[alan123hk]

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.

 

[WWGD]

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.

 

[haruspex]

One possibility would be to use logarithms.

How about the binomial expansion?