Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given,show that the area of the triangle is maximum when the angle between them is $\Large \frac{1}{3}$

Can you answer this question?

1 Answer

0 votes
  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $ f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii)By pythogorus theorem , Find $f''(x)$ and check the value of $ f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum of that point.
Step 1
Let $ABC$ be a right triangle with base = $x$ and hypotenuse $AB=y$
It is given that
$ x+y=k$
Let $ \theta$ be the angle between the base and the hypotenuse.
Area of the triangle $ABC$ is
$A=\large\frac{1}{2}\times BC \times AC$
We know $BC =x \: and \: AC = \sqrt{y^2-x^2}$
Now substitute for $BC\: and \: AC$,
$ \therefore A = \large\frac{1}{2} \times x \times \sqrt{y^2-x^2}$
Squaring on both sides we get,
But $x+y=k \Rightarrow y=k-x$
Substituting for $y$ we get,
$A^2= \large\frac{x^2}{4} [(k-x)^2-x^2]$
$ = \large\frac{x^2}{4} [ k^2+x^2-2kx-x^2]$
$ \large\frac{x^2}{4} [ k^2-2kx]$
$ \therefore A^2 = \large\frac{x^2k^2-2kx^3}{4}$
Step 2
Now differentiating w.r.t $x$ we get,
$ 2A.\large\frac{dA}{dx} = \large\frac{1}{4} [2k^2x-6kx^2]$
$ \Rightarrow \large\frac{dA}{dx} = \large\frac{1}{8A} \times 2 (k^2x-3kx^2)$
$ = \large\frac{1}{4A} ( k^2x-3kx^2)$
For maximum or minimum, we must have
$ \large\frac{dA}{dx}=0$
$ \Rightarrow \large\frac{1}{4A}(k^2x-3kx^2)=0$
$ \Rightarrow k^2x-3kx^2=0$
$ x(k^2-3kx)=0$
$ \Rightarrow k^2-3kx=0$
$ k^2=3kx$
$ \Rightarrow x=\large\frac{k^2}{3k}$
Step 3
Now differentiate $ \large\frac{dA}{dx}$ again w.r.t $x$
$ \large\frac{d^2A}{dx^2}=\large\frac{1}{4A}[k^2-6kx]$
Substituting for $ x$ we get,
$ \large\frac{d^2A}{dx^2}=\large\frac{1}{4A}[k^2-6k.\large\frac{k^2}{3k}]$
$ =\large\frac{1}{4A} [ k^2-2k^2]$
$ = - \large\frac{k^2}{4A} < 0$
This $A$ is maximum when $ x = \large\frac{k}{3}$
when $ x = \large\frac{k}{3}, \: y = k-\large\frac{k}{3}$
$ = \large\frac{2k}{3}$
$ \cos \theta = \Large\frac{x}{y}=\Large\frac{\Large\frac{k}{3}}{2\Large\frac{k}{3}}= \Large\frac{1}{2}$
$ \Rightarrow \theta = \large\frac{\pi}{3}$


answered Aug 6, 2013 by thanvigandhi_1
edited Aug 6, 2013 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App