Find the largest possible area of a triangle that has two sides of lengths 30 and 40.
The largest possible area will be when you have a right-triangle. The two sides are 30 and 40 and the hypotenuse is 50. The area of this triangle is therefore ....
A non-calculus proof (and I will scale the triangle down by a factor of 10 to sides of 3 and 4):
Let the third side be x. It follows from Heron's formula that:
$\displaystyle A^2 = \left( \frac{7 + x}{2} \right) \left( \frac{1 + x}{2} \right) \left( \frac{x - 1}{2} \right) \left( \frac{7 - x}{2} \right)$
$\displaystyle = - \frac{1}{16} (x + 7) (x - 7) (x + 1) (x - 1)$
$\displaystyle = - \frac{1}{16} (x^2 - 49)(x^2 - 1) = - \frac{1}{16} (x^4 - 50 x^2 + 49)$
$\displaystyle = - \frac{1}{16} \left[ (x^2 - 25)^2 - 576\right] = - \frac{1}{16} (x^2 - 25)^2 + \frac{576}{16}$.
Since the first term is always negative the maximum area will occur when $\displaystyle x^2 - 25 = 0 \Rightarrow x = 5$. Therefore you have a right-triangle.
Clearly the square of the maximum area is then $\displaystyle \frac{576}{16} = 36$ and so the maximum area is 6 (for this scaled down triangle).
For a triangle:
$\displaystyle \text{Area} =\dfrac{ \text{Base} \times \text{Height}}{2}$
Use either 30 or 40 as your Base and the other will be the radius of a circle. (See the attached sketch.)
$\displaystyle H = R \sin \theta $
$\displaystyle A = \dfrac{B \times H}{2}
$
Let theta run through the values from zero to 180 degrees. Calculate the Height of the triangle and then compute the area.
You may be able to see that the area of a triangle with theta = 2 degrees has the same area as a triangle with theta = (180 - 2 )degrees.
& if theta = 30 degrees the area is the same and theta = (180 - 30) degrees.
That should reduce the number of computations required.
At some point you will reach a maximum area.
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