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The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y=12-x^{2} such that the rectangle lies inside the parabola, is : 

  • Option 1)

    36

     

     

  • Option 2)

    32

  • Option 3)

    18\sqrt{3}

  • Option 4)

    20\sqrt{2}

Answers (1)

best_answer

 

Standard equation of parabola -

y^{2}=4ax

- wherein

 

\\\textup{Area} (A) = 2\alpha (12-\alpha^2) \\\\\frac{dA}{d\alpha} = 0 \\\\\Rightarrow 2(12 - 3\alpha^2) = 0 \Rightarrow \alpha = \pm 2 \\\\\Rightarrow A = 4(12-4) = 32


Option 1)

36

 

 

Option 2)

32

Option 3)

18\sqrt{3}

Option 4)

20\sqrt{2}

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