Question

If x \geq 0, y \geq 0 such that x^2 +\dfrac{y^2}{2} =1, find the maximum value of x\sqrt{1+y^2} .

Collected in the board: Inequality

Steven Zheng posted 7 hours ago

Answer

\because x^2 +\dfrac{y^2}{2} =1

x^2 +\dfrac{y^2+1}{2} = \dfrac{3}{2}

Transform the expression x\sqrt{1+y^2} to the form so that Arithmetic Mean and Geometric Mean Inequality could be applied

x\sqrt{1+y^2} = \sqrt{2} \sqrt{\dfrac{ x^2(1+y^2)}{2}}

\sqrt{ab} \leq \dfrac{a+b}{2}

Here a = x^2 and b = 1+y^2

Therefore,

\sqrt{2} \sqrt{\dfrac{ x^2(1+y^2)}{2}}\leq\sqrt{2} \cdotp \dfrac{x^2 + 1+y^2}{2} = \sqrt{2} \cdotp (x^2 + \dfrac{ 1+y^2}{2} - \dfrac{x^2}{2} )

=\dfrac{3\sqrt{2}}{2} - \dfrac{\sqrt{2}x^2}{2}

which is the quadratic function in terms of x

Obviously when x = 0, the maximum value of the expression x\sqrt{1+y^2} is \dfrac{3\sqrt{2}}{2}

Steven Zheng posted 6 hours ago

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