*This one requires multiple skills and tools, including calculus.
Given:

y = b  a*x2 (b>0, a>0, y>=0)

the total length of the curve is 10
Problem:
Find the value of b which maximizes the closed area between the curve and the x axis.
*This one requires multiple skills and tools, including calculus.
Given:
y = b  a*x2 (b>0, a>0, y>=0)
the total length of the curve is 10
Problem:
Find the value of b which maximizes the closed area between the curve and the x axis.
arc length
[spoiler]Well (I think) I’ve got (2 sqrt(4(ab)^2 + ab) + arcsinh(2 sqrt(ab))) / (2a) = 10 from the arc length restriction, but I have no idea how to get that in terms of either a or b. I’m pretty tired, so I’ll take another look tomorrow sometime not after midnight. Once you have a nice relationship with a as a function of b or vice versa it should be easy to optimize. Maybe a nice function doesn’t exist though, not sure.
Also, it’s pretty easy to get a as a function of ab, I’ll have to think about if that’s actually helpful though. Might be able to get the inverse of it.[/spoiler]
solution?
[spoiler]Ok, next I defined a variable c=ab, and got a as a function of c using the arc length. I also got area as a function of c. Then, because the derivative of the area is some insanely long expression, I put it into my CAS and got the answer ab = 1.515, a=.817 and b=1.853.[/spoiler]
EDIT: somewhere I lost a factor of two, the actual answer is:
actual solution
[spoiler]a=.626, b=4.06[/spoiler]
Plug those numbers into your arc length expression and see if you get “10”.
Plug those numbers into your area expression and tell me what you get.
area
[spoiler]12.927. Also noticed that, approximately, a*sqrt(a) = 1/b * sqrt(b). Not sure if that’s a coincidence or if figuring that out gives an easier way to solve the problem.[/spoiler]
Doublecheck your area formula.
If you’re sure it’s correct, please post it.
Nope, it was wrong. Correct area, a and b are
answer
[spoiler]a=.285, b=3.34, area = 15.2[/spoiler]
Nice job.
Want some reps? Please post your work and explain each step.
Here’s how I did it: https://docs.google.com/document/d/1bdR0C8U9UqU1sqlJego0XTq0KU2q4AaXTiEgMtOkw/edit?usp=sharing
Very interesting approach.
On page2 of the Google doc you say “Assume dR/dc=0, Solve for c”… But I don’t see dR/dc anywhere, and I don’t see you setting dR/dc=0 and solving analytically for c.
It looks like you gave a©, b©, and R© to desmos, and let it figure out dR/dc (numerically?) and find the zero crossing to get the desired value of c. Yes? I’m not familiar with the desmos calculator so I’m guessing.
If you would add just a bit more explanation to the Google doc for the benefit of future readers that would be most helpful.
That’s correct, I did it numerically with the calculator. The Google doc now clarifies this. I’ll look into an analytic solution.
It may not exist. My CAS choked on it.
Using your R(c) method, a numerical root finder quickly converges to a precise value for c.
But you can get the same result without differentiating R(c), by simply plotting R(c) and locating the extremum.
*
I get 3 arcsinh(2 sqrt©) * sqrt(4c^2+c) = 2c+8c^2. I doubt there is a nice closed form of this.
*For completeness, here’s how to solve the problem using constrained nonlinear optimization.
It’s a bit easier to set up. All you need is the length and area as a function of a and b.
The area is the objective (to be maximized), and the length is a constraint on the values of a and b.
*
Any college students out there taking a course in numerical methods?