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Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 34 maths textbook solution.

Answers (1)

(2, 2)

Hint: For maximum or minimum value of f, we must be have f'(x) =0

Given:

y^2=2x

x=\frac{y^2}{2}                     ........(1)

d^2= (x-1)^2+(y-4)^2 using the distance formula

Solution:

Now,

z=d^2= (x-1)^2+(y-4)^2

z=\left ( \frac{y^2}{2}-1 \right )^2+(y-4)^2                ........ from (1)

=\frac{y^4}{4}+1-y^2+y^2+16-8y

\frac{dz}{dy}=y^3-8

\frac{dz}{dy}=0

y^3-8=0

y^3=8

y=2

Substituting the value of y in eqn (1)

x = 2

\frac{d^2z}{dy^2}=3y^2

=12>0

The required nearest point is (2,2)

 

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infoexpert24

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