Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Find the sides of a rectangle of greatest area that can be inscribed in the ellipse $ x^2+4y^2=16$

Can you answer this question?

1 Answer

0 votes
  • For greatest area, maximise area.
  • Working rule for maxima minima.
  • Find the function to be maximised or minimised
  • Convert the function into single variable
  • Apply the conditions for maximum or minimum i.e., first derivative = 0 ( w.r.to the variable converted) and second derivative positive for minima and negative for maxima.
  • Answer the question.
Step 1
Area of rectangle $ A = 2x.2y$
$ x^2=16-4y^2 \Rightarrow A^2=256y^2-64y^4$
Step 2
$A^2 = (4xy)^2 = 16x^2y^2$
$ A^2 = 16y^2(16-4y^2)$
$ 256y^2-64y^4$
Step 3
$ \Rightarrow y = \sqrt 2$
$ x = 2\sqrt 2$
Step 4
$ \large\frac{d^2}{dy^2}(A^2)=$ -ve $\Rightarrow A$ is maximum.
The sides of rectangle are
$ 4\sqrt 2$ and $ 2\sqrt 2$
answered Apr 12, 2013 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App