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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 43.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 43.

Answers (1)

S =9

Hint: For maximum or minimum value of S, we must have \frac{dS}{dD}=0

Given:

The equation of line passing through (1,4) with slope m is given by 

y-4=m(x-1)                     ........(1)

Solution:

Sub (y=0) value in eqn (1)

0-4=m(x-1)

\frac{-4}{m}=x-1

x=\frac{m-4}{m}

Sub (x=0) value in eqn (1)

y-4=m (0-1)

y=-m+4

x=-(m-4)

So the intercepts coordinates axes are \frac{m-4}{m},-(m-4)

\begin{aligned} &S=\frac{m-4}{m}-(m-4) \\ &\frac{d S}{d m}=\frac{4}{m^{2}}-1 \\ &\frac{d S}{d m}=0 \\ &\frac{4}{m^{2}}-1=0 \\ &\frac{4}{m^{2}}=1 \\ &m^{2}=4, m=\pm 2 \end{aligned}

\begin{aligned} &\frac{d^{2} S}{d m^{2}}=\frac{-8}{m^{3}} \\ &\left(\frac{d^{2} S}{d m^{2}}\right)_{m=2}=\frac{-8}{2^{3}}=-1<0 \end{aligned}

Sum is minimum at m = 2

\left(\frac{d^{2} S}{d m^{2}}\right)_{m=-2}=\frac{-8}{(-2)^{3}}=1>0

So the sum is maximum at m = -2

Thus,

\mathrm{S}=\left[\frac{-2-4}{-2}-(-2-4)\right]=3+6=9

 

 

Posted by

infoexpert24

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