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.
I decided not to use Heron's formula. Instead I solved graphically with Desmos and then turned to Wolfram near the end.
https://www.desmos.com/calculator/etplkxj1ej
Since a=2, plot points at (0,0) and (2,0)
Then make circles of radii c and 4-2c at either end.
The triangle inequality can be used to show 2/3 < c < 2 so I made a slider for c with these bounds.
The y coordinate of the intersection of these points is then the area of the triangle.
The x-coordinate is solved in eq.6 and the y in eq.8
Leaving off the radical, ask wolfram|alpha for the local maximum in the range.
It says when c=3-sqrt(11/3) the max is (69-11sqrt(33))/6
The area is the square root of this value.
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Posted by Jer
on 2023-09-04 16:52:52 |
exact solution
