# [(a-b)2 + (a2/20 - { (17-b)x(b-13) }1/2 )2]a, b are variables (like usually used x.)can you tell me how to find the minimum value of the above expression and for what values of a and b will it be maximum.........please help....

Jitender Singh IIT Delhi
8 years ago
Ans:
$f(a, b) = (a-b)^{2}+(\frac{a^{2}}{20}-\sqrt{(17-b).(b-13)})^{2}$
$f(a, b) = a^{2} + \frac{a^{4}}{400}-2ab-\frac{a^{2}}{10}.\sqrt{(17-b).(b-13)}+ ((17-b).(b-13))$$13\leq b\leq 17$
$\frac{\partial f(a, b)}{\partial a} =2a + \frac{a^{3}}{100}-2b-\frac{a}{5}\sqrt{(17-b)(b-13)} = 0$
From here, you will get the value of a in the form of b.
Put that value in the equation & you will get the equation in b.
Then differentiate the function w.r.t b & equate it to zero.
Calculate b from the equation & select appropriate from the value b/w 13 & 17 for maxima & minima.
Thanks & Regards
Jitender Singh
IIT Delhi