Question

If one side of a triangle has length \(a\) and another has length

\(2a\), show that the largest possible area of the triangle is \({a^2}\).

Short answer

It is proved that the largest possible area of the triangle is \({a^2}\).

Step-by-step solution

Application of Derivative

We can apply the derivative of a function to find the closest point of a curve to a point or the furthest point of a curve to another point. We find these points with the concept of local minimaand local maximarespectively.

We find local minima and maxima of a function by finding critical pointsand the second derivativeof the function.

Critical points are the \(x\) values that satisfy \(f'\left( x \right) = 0\).

The function has a local maximum at a point \(x\) if \(f''\left( x \right) < 0\).

The function has a local minimum at a point \(x\) if \(f''\left( x \right) > 0\).

Finding equations related to a rectangle and the equilateral triangle

Let the one interior angle of the triangle is \(\theta \).

Then the height of the triangle will be \(h = 2a\sin \theta ,0 < \theta < \pi \).

Hence the area of the triangle will be \(A = \frac{1}{2}a\left( {2a} \right)\sin \theta \Rightarrow A = {a^2}\sin \theta \).

Finding the Local Maxima of the area function

Differentiating we get,

\(A'\left( \theta \right) = {a^2}\cos \theta \)

So the critical point will be,

\(\begin{aligned}{c}A'\left( x \right) = 0\\\cos \theta = 0\\\theta = \frac{\pi }{2}\end{aligned}\)

Hence the maximum value of the function \(A\) will be at the point \(\theta = \frac{\pi }{2}\).

Thus the largest area of the triangle is \(A = {a^2}\sin \frac{\pi }{2} \Rightarrow A = {a^2}\).

Hence proved.