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.
(In reply to
Possible methodology for exact solution by Larry)
I finished what I started in my earlier post, same solution as Kenny M.
With side=2 as the base and h as the altitude, the other sides are b and c, but put b in terms of c since the problem asks for c. So the sides are 2, c, and 4-2c.
x^2 + h^2 = (4-2c)^2
(2-x)^2 + h^2 = c^2
subtracting:
x = (1/4)(3c^2 - 16c + 20)
substitute into first equation:
16h^2 = -9c^4 + 96c^2 - 312c^2 + 384c -144
Take derivative:
32h*(dh/dc) = -36c^3 + 288c^2 - 624c + 384 (c=2 is a factor)
32h*(dh/dc) = (c-2)(-36c^2 + 216c - 192)
(-36c^2 + 216c - 192) = 0
c = 3 ± √33/3 reject the + version since c is part of a rope and the entire rope is 4.
Solution for c: c = 3 - √33/3
c = 1.085145784487323780
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Posted by Larry
on 2023-09-07 08:46:33 |