If y=a+bx3/2x5/4 and y′=0 at x=5then the value of a4b4is

Solution:

$y = \frac{a + {bx}^{3 / 2}}{x^{5 / 4}}$
or $y^{'} = \frac{\frac{3}{2} {bx}^{1 / 2} x^{5 / 4} - \frac{5}{4} x^{1 / 4} (a + {bx}^{3 / 2})}{x^{5 / 2}}$
according to the question,
$0 = \frac{\frac{3}{2} \times b 5^{1 / 2} \times 5^{5 / 4} - \frac{5}{4} \times 5^{1 / 4} (a + b \times 5^{3 / 2})}{5^{5 / 2}}$
or $\frac{3 b}{2} \times 5^{7 / 4} - a \times \frac{5^{5 / 4}}{4} - 5 b \times \frac{5^{7 / 4}}{4} = 0$
or $b \times 5^{7 / 4} = a \times 5^{5 / 4}$
or $b \sqrt{5} = a$
or $a : b = \sqrt{5} : 1$

$\Rightarrow \frac{a^{4}}{b^{4}} = \frac{25}{1} = 25$

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