Math help desperately needed!
Postmaster
Joined: Aug 2004
Posts: 4,004
Likes: 1
From: North of Boston MA
Since the formula given describes that of a parabola, there is only one solution where the slope of the derivative equals 0.
So, the derivative of the equation, when set equal to zero looks like this:
20-s=0
s=20
Enter 20 back into the original equation and P=430
So, the derivative of the equation, when set equal to zero looks like this:
20-s=0
s=20
Enter 20 back into the original equation and P=430
Cross-Country
Joined: Aug 2000
Posts: 61
Likes: 1
From: Cincinnati, OH
No derivatives...no calculus....just algebra. That's what he said. Here's what you do. Since it is a second order, it is a parabola, as so many have pointed out. Since the squared term is negative, the parabola has the vertex up. The parabola crosses the X axis at two points. Half-way between these two points will be the X coordinate of the vertex, which is the 's' that you are trying to find. Set the equation to 0 (0 = -0.5s^2+20s+230). Solve by quadratic (you know....b^2 +/- squart root of.....etc etc). This will give you your two X axis points where the curve crosses the X axis (one will be + the other - for this equation). Add those two X values together, divide by 2 and that is your 'S'. Stick this 'S' back into your equation to get your 'P(S)'. S = 20. P(S) = 430. Done! Simple....heh!
Last edited by lo; Apr 10, 2007 at 01:04 PM.
Cargo Master
Joined: Dec 2006
Posts: 3,489
Likes: 45
From: Central NY
Originally Posted by Caplax40
Thank you for your help!!! I don't know how you went finding an answer of 20 other than trial and error. Using those two numbers I was able to figure out what transformation model to use.
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