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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 30 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 30 maths textbook solution.

Answers (1)

(4,-4)

Hint: For maximum or minimum value of z must have \frac{dz}{dy}=0

Given: y^2=4x

x=\frac{y^2}{4}               ........(1)

Also,

d^2=(x-2)^2+(y+8)^2 using distance formula

Solution:

Now,

z=d^2=(x-2)^2+(y+8)^2

z=\left ( \frac{y^2}{4}-2 \right )^2+(y+8)^2             ...... from (1)

\begin{aligned} &=\frac{y^{4}}{16}+4-y^{2}+y^{2}+54+16 y \\ &\frac{d z}{d y}=\frac{4 y^{3}}{16}+16 \\ &\frac{d z}{d y}=0 \\ &\frac{4 y^{3}}{16}+16=0 \end{aligned}

4y^3=-64, y=-4

Substituting the value of y in equation (1)

x =4

\frac{d^{2} z}{d y^{2}}=\frac{12 y^{2}}{16}=12>0

The required nearest point is (4,-4)

 

 

Posted by

infoexpert24

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