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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the sides of a rectangle of greatest area that can be inscribed in the ellipse $ x^2+4y^2=16$

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Toolbox:
  • 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
$\large\frac{d}{dy}(A^2)=512y-256y^3-0$
$ \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
 

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