Name: Increasing Functions
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: The GDP of India is either increasing or decreasing, depending on a variety of factors. Let’s understand what an increasing function is, by using derivatives...

Question

Show that the right triangle of maximum area that can be inscribed in a circle is an isosceles triangle.

Accepted Answer

Given that a right angle triangle is inscribed inside the circle.

Let the radius of the circle is $r$ .

Let $x$ and $y$ be the base and height of the right angle triangle $∆ A B C$ as shown in the diagram.

$∴ A B^{2} = A C^{2} + B C^{2}$

We have $A B = 2 r, A C = y$ and $B C = x$

Hence,

$4 r^{2} = x^{2} + y^{2}$

$\Rightarrow y = \sqrt{4 r^{2} - x^{2}} . . . . . (i)$

Now, Area of the $△ A B C$ is

$A = \frac{1}{2} \times base \times height$

$\Rightarrow A = \frac{1}{2} \times x \times y$

Now substituting $(i)$ in the area of the triangle,

$A = \frac{1}{2} x (\sqrt{4 r^{2} - x^{2}})$

[Squaring both sides]

$Z = A^{2} = \frac{1}{4} x^{2} (4 r^{2} - x^{2}) . . . (ii)$

Differentiating the equation $(ii)$ with respect to $x$ :

$\frac{d Z}{d x} = \frac{d [\frac{1}{4} x^{2} (4 r^{2} - x^{2})]}{d x}$

$\Rightarrow \frac{d Z}{d x} = \frac{1}{4} [(4 r^{2} - x^{2}) \frac{d (x^{2})}{d x} + x^{2} \frac{d (4 r^{2} - x^{2})}{d x}]$

$\Rightarrow \frac{d Z}{d x} = \frac{1}{4} [(4 r^{2} - x^{2}) (2 x) + x^{2} (- 2 x)]$

$∴ \frac{d Z}{d x} = \frac{1}{4} [8 r^{2} x - 2 x^{3} - 2 x^{3}]$ $= [2 r^{2} x - x^{3}] . . . (i i i)$

Now, $\frac{d Z}{d x} = 0$

$\Rightarrow \frac{d Z}{d x} = 2 r^{2} x - x^{3} = 0$

$\Rightarrow 2 r^{2} x = x^{3}$

$\Rightarrow x^{2} = 2 r^{2}$

$\Rightarrow x = \pm \sqrt{2 r^{2}}$

As $x$ is the base of triangle, $x = - r \sqrt{2}$ is rejected

Differentiating Equation $(iii)$ with respect to $x$ again:

$\frac{d^{2} Z}{d x^{2}} = \frac{d [2 r^{2} x - x^{3}]}{d x}$

$\Rightarrow \frac{d^{2} Z}{d x^{2}} = \frac{d [2 r^{2} x]}{d x} + \frac{d [- x^{3}]}{d x}$

$\Rightarrow \frac{d^{2} Z}{d x^{2}} = 2 r^{2} - 3 x^{2} . . . (iv)$

By Second derivative test,

${[\frac{d^{2} Z}{d x^{2}}]}_{x = r \sqrt{2}} = 2 r^{2} - 3 {(r \sqrt{2})}^{2} = 2 r^{2} - 6 r^{2} = - 4 r^{2} < 0$

So, the function $A$ is maximum at $x = r \sqrt{2}$ .

Now substituting $x = r \sqrt{2}$ in equation $(i)$ :

$y = \sqrt{4 r^{2} - (r \sqrt{2})^{2}} = \sqrt{4 r^{2} - 2 r^{2}} = r \sqrt{2}$

As $x = y = r \sqrt{2}$ , the base and height of the triangle are equal, which means that two sides of a right angled triangle are equal.

Hence the given triangle, which is inscribed in a circle, is an isosceles triangle with sides $A C$ and $B C$ equal.

Show that the right triangle of maximum area that can be inscribed in a circle is an isosceles triangle.

Important Questions on Applications of Derivatives

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