Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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)

 

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question