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

Using calculus, prove that the perpendicular distance of the line

3 x + 4 y + 5 = 0

from the origin is 1 unit.

Accepted Answer

We have to prove that the perpendicular distance of the line $3 x + 4 y + 5 = 0$ from the origin is $1$ unit.

Any point on the given line is of the form $(x, \frac{- 5 - 3 x}{4})$ .

So, the distance between any point on the line from origin is given by $l = \sqrt{x^{2} + y^{2}} = \sqrt{x^{2} + {(\frac{- 5 - 3 x}{4})}^{2}}$ .

Perpendicular distance of the given line from origin is the minimum of distance $l$ .

We have, $l = \sqrt{x^{2} + {(\frac{- 5 - 3 x}{4})}^{2}}$

$\Rightarrow l^{2} = x^{2} + {(\frac{- 5 - 3 x}{4})}^{2}$

$\Rightarrow l^{2} = x^{2} + \frac{{(- 5)}^{2} + 30 x + 9 x^{2}}{16}$

$\Rightarrow 16 l^{2} = 25 x^{2} + 30 x + 25$

Differentiating above equation w.r.t. $x$ , we get

$16 \times 2 l \frac{d l}{d x} = 50 x + 30 . . (1)$

Now, $\frac{d l}{d x} = 0 \Rightarrow 50 x + 30 = 0$

$\Rightarrow x = \frac{- 3}{5}$

Differentiating $(1)$ w.r.t. $x$ , we get

$32 {(\frac{d l}{d x})}^{2} + 32 l (\frac{d^{2} l}{d x^{2}}) = 50$

When $\frac{d l}{d x} = 0 and x = - \frac{3}{5}$ , we get $32 {(0)}^{2} + 32 l (\frac{d^{2} l}{d x^{2}}) = 50$ i.e., $\frac{d^{2} l}{d x^{2}} = \frac{50}{32 l} > 0$ ( $l > 0$ at $x = - \frac{3}{5}$ )

Hence, $l$ is minimum when $x = - \frac{3}{5}$ .

$l = \sqrt{{(- \frac{3}{5})}^{2} + {(\frac{- 5 + \frac{9}{5}}{4})}^{2}} = \sqrt{\frac{9}{25} + \frac{16}{25}} = 1$

Thus, the perpendicular distance of the line $3 x + 4 y + 5 = 0$ from the origin is $1$ unit.

Using calculus, prove that the perpendicular distance of the line $3 x + 4 y + 5 = 0$ from the origin is 1 unit.

Important Questions on Maxima and Minima

Learn and Prepare for any exam you want !

Using calculus, prove that the perpendicular distance of the line 3x+4y+5=0 from the origin is 1 unit.

Important Questions on Maxima and Minima

Learn and Prepare for any exam you want !

Using calculus, prove that the perpendicular distance of the line $3 x + 4 y + 5 = 0$ from the origin is 1 unit.