If the sum of the lengths of hypotenuse and a side of a right angled triangle is given, then the area of the triangle is maximum when angle between them is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \pi/3 \)
Let the given side be x and the hypotenuse h, with \( x+h=k \) (a constant). The third side is
\( \displaystyle \sqrt{h^2-x^2}=\sqrt{(k-x)^2-x^2} \)
\( \displaystyle =\sqrt{k^2-2kx} \)
The area is \( A=\frac{1}{2}x\sqrt{k^2-2kx} \). It is easier to maximise \( S=A^2 \):
\( \displaystyle S=\frac{1}{4}x^2(k^2-2kx) \)
\( \displaystyle =\frac{1}{4}(k^2x^2-2kx^3) \)
\( \displaystyle \frac{dS}{dx}=\frac{1}{4}(2k^2x-6kx^2) \)
\( \displaystyle =\frac{kx}{2}(k-3x) \)
\( \displaystyle \frac{dS}{dx}=0\ \Rightarrow\ x=\frac{k}{3} \)
\( \displaystyle \frac{d^2S}{dx^2}=\frac{1}{4}(2k^2-12kx) \)
\( \displaystyle =\frac{1}{4}(2k^2-4k^2) \)
\( \displaystyle =-\frac{k^2}{2}\lt 0\ \text{at}\ x=\frac{k}{3} \)
So the area is maximum when \( x=\frac{k}{3} \), and then \( h=k-\frac{k}{3}=\frac{2k}{3} \). If \( \theta \) is the angle between the side and the hypotenuse,
\( \displaystyle \cos\theta=\frac{x}{h} \)
\( \displaystyle =\frac{k/3}{2k/3} \)
\( \displaystyle =\frac{1}{2} \)
\( \displaystyle \theta=\frac{\pi}{3} \)
Tip: This is a standard result worth remembering: when hypotenuse + side is fixed, the area is maximum when the hypotenuse is twice that side, i.e. the angle between them is \( \frac{\pi}{3} \).
NCERT Exemplar:
Application of Derivatives
