The perimeter of a right angled triangle of given hypotenuse ‘h’ is maximum if the ratio of other two sides of the triangle is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1 : 1
Let one acute angle be \( \theta \). Then the other two sides are \( h\cos\theta \) and \( h\sin\theta \), and
\( \displaystyle P=h+h\cos\theta+h\sin\theta,\quad 0\lt\theta\lt\frac{\pi}{2} \)
\( \displaystyle \frac{dP}{d\theta}=h(\cos\theta-\sin\theta) \)
Setting \( \frac{dP}{d\theta}=0 \):
\( \displaystyle \tan\theta=1\ \Rightarrow\ \theta=\frac{\pi}{4} \)
\( \displaystyle \frac{d^2P}{d\theta^2}=-h(\cos\theta+\sin\theta) \)
\( \displaystyle =-h\sqrt{2}\lt 0\ \text{at}\ \theta=\frac{\pi}{4} \)
So P is maximum at \( \theta=\frac{\pi}{4} \), where the two sides are equal: \( h\cos\frac{\pi}{4}=h\sin\frac{\pi}{4}=\frac{h}{\sqrt{2}} \). The ratio is 1 : 1.
Shortcut: With sides a and b, \( a^2+b^2=h^2 \) is fixed, and \( (a+b)^2=2(a^2+b^2)-(a-b)^2\le 2h^2 \). So a + b is largest when a = b.
NCERT Exemplar:
Application of Derivatives
