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.
Heron's formula states that given the semi-perimeter of a triangle s, in this case s=(a+b+c)/2, then the area is
A=SQRT[s*(s-a)*(s-b)*(s-c)].
Given a=2, b=b, and c=(2-b/2), then after algebra, s=(2+b/4).
We need to maximize the area, but in this case that is the same as maximizing the term under the radical, namely s*(s-a)*(s-b)*(s-c). Lets call this R.
Using what we already have for s, a, b, and c, and with more algebra, R=12*(b^2)/16-12*(b^3)/64-9*(b^4)/256. dR/db=24*b/16-36*(b^2)/64-36*(b^3)/256.
At this point, I went to the computer and found that dR/db=0 when b~1.82972 giving c~1.08514, and an area of 0.98402. This is indeed a maximum for the area of the resulting triangle, which I see now is in agreement with Charlie's previous post.
Edited on September 4, 2023, 12:54 pm
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Posted by Kenny M
on 2023-09-04 12:53:06 |
Semi-analytical solution
