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IIT JEE Entrance Exam

if a+2b+c=4,
then find the max. & min. value for the expression ab+bc+ca.
where a, b & c are real nos.

Profile image of GOPAL GUPTA
11 Years agoGrade
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1 Answer

Profile image of Nishant Vora
11 Years ago
here a+2b+c=4
=> b = 2 - \frac{a}{2} - \frac{c}{2} ...........(1)

using eqn (1) in above
ab+bc+ca
=a(2 - \frac{a}{2} - \frac{c}{2}) + c(2 - \frac{a}{2} - \frac{c}{2}) + ac
=2a - \frac{a^2}{2} - \frac{ac}{2} + 2c - \frac{ac}{2} - \frac{c^2}{2} + ac
=> 2a - \frac{a^2}{2} + 2c - \frac{c^2}{2}
=> -\frac{1}{2}[-4a + a^2 - 4c + c^2 ]
=> -\frac{1}{2}[(a^2 -4a +4) + (c^2- 4c +4) - 8 ]
=> -\frac{1}{2}[(a-2)^2 + (c-2)^2 - 8 ]
=> 4 - \frac{(a-2)^2}{2} - \frac{(c-2)^2}{2}
=> 4 - \frac{(a-2)^2}{2} - \frac{(c-2)^2}{2} \leq 4

so max value = 4


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