
Elite Member
Originally Posted by
JacDevo
What did you find?
What did you get, when you solved the system of equations with which you were provided?

Elite Member
Originally Posted by
JacDevo
What did you find?
This is a help site, not an answer site. I am not even sure I guessed correctly what the problem is. It seems plausible that the maximum distance of 120 is correct, but you neither gave that information nor confirmed what was a guess. Furthermore, it is not clear to me that the problem permits you to place the origin at point A though that makes the equation of the line very simple and certainly simplifies solving for the coefficients of the cubic.
This problem consists of several subproblems. The first is finding the coeffients of a cubic. It won't help you on a test if you are reliant on me or a cubic calculator to find such coefficients.
If we are allowed to choose point A as the origin, then we have the equation
[tex]f(0) = a(0^3) + b(0^2) + c(0) + d = 0 \implies 0 + 0 + 0 + d =0 \implies[/tex]
[tex]d = 0 \implies f(x) = ax^3 + bx^2 + cx = x(ax^2 + bx + c).[/tex]
Now quite clearly your cubic has d = 120, which certainly follows from making f(0) the maximum, but makes for a very different kind of cubic. Consequently, if you are required to place the maximum at the origin, you need to revise all the equations I provided. If however you are free to use A as the origin, you may find the arithmetic easier and will definitely find that the equation of the line is much easier to work with.
Assuming that you are allowed to set A as the origin and that you see why d = 0, you can find the remaining coefficients by solving the following system of three linear equation in three unknowns
[tex]0 = 3a(30^2) + 2b(30) + c[/tex]
[tex]120 = 30\{a(30^2) + b(30) + c\} \implies 4 = 900a + 30b + c[/tex]
[tex]0 = 90\{ a(90^2) + b(90) + c\} \implies 0 = 8100a + 90b + c[/tex]

Elite Member
It almost seems a waste of time to respond.
You have been told that you can find the exact equation of the cubic. You can do that either using the coordinate system you selected or the one I recommended as probably easier. I then GAVE you the system of LINEAR equations to do that using my recommended coordinate system. You asked no questions about how that system was found so I presume you can develop your own equations for your preferred coordinate system.
Now your cubic calculator may well have computed the proper area, and all will be well on your test if you have access to such a calculator.
It is true that you will eventually get quartic equations for the area. But you will not have to find the roots of a quartic to find k.
If C(x) is the equation of the cubic and L(x) of the line you must equate the integrals of
C(x)  L(x) and of L(x)  C(x) over intervals determined by the coordinate system. But if you think about it you should realize that the fourth and second powers of k will drop out of the equation.
Finally you have made life very difficult for yourself by using decimal approximations instead of exact fractions.
I shall proceed only in one of two ways. Either you develop an exact equation for the cubic using your coordinate system (probably best), or you complete the process for finding the exact equation under my recommended coordinate system. Once that is under our belts, we can talk about finding k. Otherwise, I am done.
Last edited by JeffM; 04142017 at 10:42 AM.
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