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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 32 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 32 maths textbook solution.

Answers (1)

(\pm 2\sqrt{2},4)

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=2y nearest (0,5)

x^2=2y

y=\frac{x^2}{2}                    .........(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}{2}-5 \right )^2                 ....... from (1)

\begin{aligned} &=x^{2}+\frac{x^{4}}{4}+25-5 x^{2} \\ &\frac{d z}{d y}=2 x+x^{3}-10 x \\ &\frac{d z}{d y}=0 \\ &x^{3}-8 x=0 \\ &x^{2}=8, x=\pm 2 \sqrt{2} \end{aligned}

Substituting the value of x in eqn (1)

y =4

\frac{d^2z}{dy^2}=3x^2-8

=24-8=16>0

The required nearest point (\pm 2\sqrt{2},4)

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