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if x is real ,then maximum value of y=2(a-x)(x+whole root(x^2+b^2))

if x is real ,then maximum value of


  y=2(a-x)(x+whole root(x^2+b^2))

Grade:12th Pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find answer to your question below

y = 2(a-x)(x+\sqrt{x^2+b^2})
For maxima and minima,
y' = 2(a-x)(1+\frac{1}{\sqrt{x^2+b^2}})+(-2)(x+\sqrt{x^2+b^2})
y’ = 0
2(a-x)(1+\frac{1}{\sqrt{x^2+b^2}})+(-2)(x+\sqrt{x^2+b^2}) = 0
2(a-x)(x+\sqrt{x^2+b^2})=2(x+\sqrt{x^2+b^2}).\sqrt{x^2+b^{2}}
(a-x)=\sqrt{x^2+b^{2}}
a^2+x^2-2ax=x^2+b^{2}
x_{0} = \frac{a^2-b^2}{2a}

[y]_{max} = 2(a-x_{0})(x_{0}+\sqrt{x_{0}^2+b^2})
[y]_{max} = 2(a-\frac{a^2-b^2}{2a})(\frac{a^2-b^2}{2a}+\sqrt{(\frac{a^2-b^2}{2a})^2+b^2}
[y]_{max} = \frac{a^2+b^2}{a}(\frac{a^2-b^2}{2a}+\frac{a^2+b^2}{2a})
[y]_{max} = a^2+b^{2}

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