Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

A tangent to the curve

y = 1 - x^{2}

is drawn so that the abscissa

x_{0}

of the point of tangency belongs to the interval

(0, 1] .

The tangent at

x_{0}

meets the

x

-axis and

y

-axis at

A & B

respectively, then find the minimum area of the triangle

O A B,

where

O

is the origin.

Accepted Answer

Given a curve $y = 1 - x^{2}$

$\Rightarrow \frac{d y}{d x} = - 2 x$

The equation of tangent on the curve at abscissa $x_{0}$ is $y - (1 - x_{0}^{2}) = - 2 x_{0} (x - x_{0})$

$\Rightarrow y = - 2 x_{0} x + x_{0}^{2} + 1$

Given, the tangent at $x_{0}$ meets the $x$ -axis and $y$ -axis at $A & B$ respectively.

$\Rightarrow A = (\frac{x_{0}^{2} + 1}{2 x_{0}}, 0) & B = (0, x_{0}^{2} + 1)$

Let, $∆$ be the area of the triangle $O A B$ , where $O$ is the origin.

$\Rightarrow ∆ = \frac{1}{2} |\begin{matrix} 0 & 0 & 1 \\ \frac{x_{0}^{2} + 1}{2 x_{0}} & 0 & 1 \\ 0 & x_{0}^{2} + 1 & 1 \end{matrix}|$

$\Rightarrow ∆ = \frac{1}{4} \frac{{(x_{0}^{2} + 1)}^{2}}{x_{0}}$ Sq. units $. . . . . (i)$

$\Rightarrow \frac{d ∆}{d x_{0}} = \frac{1}{4} \frac{2 (x_{0}^{2} + 1) 2 x_{0} x_{0} - 1 \times {(x_{0}^{2} + 1)}^{2}}{x_{0}^{2}}$

$\Rightarrow \frac{d ∆}{d x_{0}} = \frac{3 (x_{0}^{2} + 1)}{4 x_{0}^{2}} (x_{0} + \frac{1}{\sqrt{3}}) (x_{0} - \frac{1}{\sqrt{3}})$

Given, $x_{0} \in (0, 1]$

So, $\frac{d ∆}{d x_{0}}$ changes its sign from negative to positive about $x_{0} = \frac{1}{\sqrt{3}}$ i.e. $x_{0} = \frac{1}{\sqrt{3}}$ is the point of local minima.

Put, $x_{0} = \frac{1}{\sqrt{3}}$ in the equation $(i)$ .

$\Rightarrow ∆_{\min} = \frac{1}{4} \frac{{(\frac{1}{3} + 1)}^{2}}{\frac{1}{\sqrt{3}}} =$ $\frac{4 \sqrt{3}}{9}$ sq. units.

A tangent to the curve $y = 1 - x^{2}$ is drawn so that the abscissa $x_{0}$ of the point of tangency belongs to the interval $(0, 1] .$ The tangent at $x_{0}$ meets the $x$ -axis and $y$ -axis at $A & B$ respectively, then find the minimum area of the triangle $O A B,$ where $O$ is the origin.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

A tangent to the curve y=1-x2 is drawn so that the abscissa x0 of the point of tangency belongs to the interval (0, 1]. The tangent at x0 meets the x-axis and y-axis at A & B respectively, then find the minimum area of the triangle OAB, where O is the origin.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

A tangent to the curve $y = 1 - x^{2}$ is drawn so that the abscissa $x_{0}$ of the point of tangency belongs to the interval $(0, 1] .$ The tangent at $x_{0}$ meets the $x$ -axis and $y$ -axis at $A & B$ respectively, then find the minimum area of the triangle $O A B,$ where $O$ is the origin.