Question

. If  the sum  of the lengths  of the hypotenuse  and a side of a right  triangle is given, show that  the area of the triangle is maximum  when the angle between them is π /3.

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Answer 1

Answer:

Step-by-step explanation:

Let that ABC is a right angle triangle having right angle B.

Let the sum of its side AB and AC is  p.

AB + AC = p

If

AB = x

AC = p - x

Using Pythagoras theorem,

(BC)² = (p - x)²  - x²

(BC)² = p² + x² -2px - x²

(BC)² = p² - 2px

(BC) =√(p²-2px)

Let the are of ΔABC be A.

A = (base × height)/2

$A=\frac{x\sqrt{x^2-2px}}{2}$

Differentiation both sides with respect to x,

$\frac{dA}{dx}=\frac{1}{2}[\sqrt{p^2-2px}+x.\frac{(-2p)}{2\sqrt{p^2-2px}}]$

$\frac{dA}{dx}=\frac{1}{2}[\sqrt{p^2-2px}-\frac{px}{\sqrt{p^2-2px}}]$

$\frac{dA}{dx}=\frac{1}{2}[\frac{p^2-2px-px}{\sqrt{p^2-2px}}]$

$\frac{dA}{dx}=\frac{1}{2}[\frac{p^2-3px}{\sqrt{p^2-2px}}]\;\;\;\;\;\;\;......i)$

For maximum and Minimum values of x,

$\frac{dA}{dx}=0\\\;\\\frac{1}{2}[\frac{p^2-3px}{\sqrt{p^2-2px}}]=0\\\;\\p^2-3px=0\\\;\\p(p-3x)=0\\\;\\p=0\;\;or\;\;p=3x$

But at p = 0 , triangle is not possible. So we will consider p = 3x only

Now, Differentiating the equation i) again with respect to x.

$\dfrac{d^2A}{dx^2}=\dfrac{1}{2}[\dfrac{\sqrt{p^2-2px}(-3p)-(p^2-3px)[\frac{-p}{\sqrt{p^2-2px}}])}{p^2-2px}]$

$\dfrac{d^2A}{dx^2}=\frac{1}{2}\dfrac{-3p(p^2-2px)+p(p^2-3px)}{(p^2-2px)^{\frac{3}{2}}}$

Putting p = 3x in above equation;

$(\frac{d^2A}{dx^2})_{p=3x}=\frac{1}{2}[\frac{-3(3x)(9x^2-6x^2)+3x(9x^2-9x^2)}{(9x^2-6x^2)^{\frac{3}{2}}}]$

$(\frac{d^2A}{dx^2})_{p=3x}=\frac{1}{2}[\frac{-9x(3x^2)}{(3x^2)^(\frac{3}{2})}]$

$(\frac{d^2A}{dx^2})_{p=3x}=-\frac{3\sqrt{3}}{2}$

Here second derivative has negative sing, it means at p = 3x area of triangle is maximum.

Now, From Image;

$\cos A=\frac{x}{3x-x}\\\;\\\cos A=\frac{1}{2}\\\;\\\cos A=\cos\frac{\pi}{3}\\\;\\A=\frac{\pi}{3}$