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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 29 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 29 maths textbook solution.

Answers (1)

(\pm 2\sqrt{3},3)

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

Given: Let the point (x,y) on the curve x^2=4y  nearest to (0,5)

x^2=4y

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

Also,

d^2=(x)^2+(y-5)^2 using the distance formula

Solution:  

Now,

z=d^2=(x)^2+(y-5)^2

z=(x)^2+\left ( \frac{x^2}{4} -5\right )^2                     ....... from (1)

\begin{aligned} &=x^{2}+\frac{x^{4}}{16}+25-\frac{5 x^{2}}{2} \\ &\frac{d z}{d y}=2 x+\frac{4 x^{3}}{16}-5 x \\ &\frac{d z}{d y}=0 \\ &2 x+\frac{4 x^{3}}{16}-5 x=0 \end{aligned}

\frac{4x^3}{16}=3x

x^3=12x

x^2=12

x=\pm 2\sqrt{3}

Substituting the value of x is equal (1)

y =3

\frac{d^{2} z}{d y^{2}}=2+\frac{12 x^{2}}{16}-5=9-3=6>0

The required nearest point is (\pm 2\sqrt{3},3)

 

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