Let ABC be a triangle with side lengths a, b, c and a=2, b+2c=4. Find the value of c which maximizes the area of the triangle.
Didn't have time to continue this until today, but thanks to Jer, I can check my work.
In my previous post, dR/db=24*b/16-36*(b^2)/64-36*(b^3)/256.
Smplifying the fractions and factoring out a "b" we get
dR/db=b*(-9/64*b^2-9/16*b+3/2).
Setting this equal to zero, we get b=0 (an obvious minimum for the areas, since the area would be zero), and we can use the quadratic formula on the level 2 polynomial that remains. Doing so, and with lots of simplification we get,
b=[9/16 +- SQRT(297/256)]/(-18/64), which further simplifies to
b=-2+-(2/3)SQRT(33). One answer is negative, which can be discarded. So b=-2+(2/3)SQRT(33) ~ 1.829708 which matches the computer based answers.
Remembering that c=2-b/2, c then simplfies to c=3-(1/3)SQRT(33)=3-sqrt(11/3) ~1.085146 , which matches what Jer found from Wolfram-Alpha analytically and the computer based answers.
I'm too tired to calculate the area, Q.E.D.
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Posted by Kenny M
on 2023-09-05 19:53:34 |
Analytical exact solution
