**Toolbox:**

- 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,

$A^2=\large\frac{x^2}{4}(y^2-x^2)$

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}$