Without using a calculator or the derivative.
Find the maximum and when it will happen. P(t)= 120t - .4t^4 +1000
Can't use wolfram or google and you have to show work.
I made the argument that its almost impossible to do this problem without the derivative or a calculator but another WSOer say it's possible but they don't know how. If someone does know how to do this I'd love to see the steps.
find the maximum when what will happen
the function?
Basically the question was asking about rabbits and it was find the point when the rabbit population is the highest.
why no deriv u noob
laplace xform alowed?
no derivative no computer paper pencil get to work.
can I do the derivative in my head and solve it
Can't give ya an exact answer, is it around 33?
just translate the polynomial in the P(t) direction so that there are degenerate roots
THE FUK IS A DEGENERATE ROOT
This is more of a consulting question than a task for quants. Here goes. Noting that we are not allowed derivatives, always look at the extremes of a function.
i) P(t=0) = 1000 clearly. At this point we note that there is no need to look at any negative values of t, since the linear function (120t) will offset any positive equivalent. ii) lim(t>inf)P(t) will clearly be negative since t^4 scales much faster than t^1, regardless of the coefficient in front.
Now, lets check for specific t values. When t= 10, a nice round number for which we know t^4, we have
iii) P(t=10) = 1200-4000+1000 0.
So already we know that the maximum must be between t = 0 and t = 10. Powers of 4 are easy to do in your head, and most people ahem should know it for values 2 to 8 or 9. Because we have the +1000 at the end for all cases, we need only consider the difference between 120t and 0.4t^4. If we say Q(t) = 120t - 0.4t^4,
iv) Q(2) = 240 - 6.4 = 233.6 v) Q(3) = 360 - 32.4 = 327.6 vi) Q(4) = 480 - 102.4 = 377.6 vii) Q(5) = 600 - 250 = 350
So the maximum value of Q(t) and therefore P(t) will occur somewhere between t = 4 and t = 5. We can then look at t = 4.5, and a decimal place around it (t = 4.4, t = 4.6). Its then a matter of how accurate you want the answer to be (in terms of t) and thereafter P(t).
+1
I'm guessing you're working on nonlinear dynamics problems (1-D at the moment it seems). You'll soon learn of how important derivatives and the Jacobian are. But this is really old school mathematics, not useful for every day life.
Interesting nonetheless.
good luck calculating 4.5...^4 in any reasonable amount of time. it's possible to solve this problem analytically without "derivatives".
4.5^4 = ((5-0.5)^2)^2 = (25-5+0.25)^2 = (20 + 1/4)^2 = (400 + 10 + 1/16) = 410.0625
[EDIT: It havnig an analytical solution means the answer should come out to a nice whole number. If we do take the derivative to check what the real solution is, we have
P'(t) = 0 = 120 - 1.6t^3. The answer to which we have t^3 = 75. This should be t = 4.2...
When you substitute this back into P(t), what do you get?]
Solutions or kindly STFUGTFO.
Taking the first derivative I get to 120^(1/3).
Without taking the derivative what I would do is ignore the 1000, and then solve this: 120t>0.4t^4, using this method I get to 480^(1/4)
P(0) = 1000 -.4t^4 and 120t offset each other to a certain extent. You want to solve for when -4t^4 completely offsets 120t and drives that sum to be negative.
So you would just be solving for .4t^4 = 120t.
.4t^3 = 120 t^3 = 300
So the answer is somewhere between 6 and 7.
No you want to maximize 120t - 0.04t^4
P(4) is greater than P(6) to P(7)
edit...p(4) sorry
Isn't the maximum found using the 2nd derivative?
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