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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 45.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 45.

Answers (1)

a = -260

Hint: For maxima and minima value of S, we must have \frac{dS}{dD}=0

Given: S= t^5-40t^3-30t^2-80t-250

Solution:  \frac{dS}{dt}= 5t^5-120t^2-60t-80

Acceleration, a=\frac{d^2S}{dt^2}= 20t^3-240t-60

\frac{da}{dt}= 60t^2-240

\frac{da}{dt}=0

60t^2-240=0

60t^2=240, t=2

Now,

\frac{d^2a}{dt^2}=120t

\frac{d^2a}{dt^2}=240>0

So, acceleration is minimum at t =2

a_{min}=20(2)^3-240(2)+60

a_{min}=160-480+60

a_{min}=-260

\therefore At =2, a = -260

Posted by

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