0 votes 0 votes Find the maximum value of f(x). n is a constant f(x)= C(x,2) *$2^{n-x}$ Calculus calculus maxima-minima engineering-mathematics + – rahul sharma 5 asked Nov 6, 2017 rahul sharma 5 892 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply just_bhavana commented Nov 6, 2017 reply Follow Share $\frac{3}{8} * 2^{n}$ ? 0 votes 0 votes rahul sharma 5 commented Nov 6, 2017 reply Follow Share yes. can you share you solution. ?I am doing some mistake and not able to find:( 0 votes 0 votes just_bhavana commented Nov 6, 2017 reply Follow Share Didn't do by naive method coz the equation turns out to be complex. Simply try for different values of x. x cannot be less than 2, it is at least 2 For x = 2, f(x) = $\binom{2}{2}*2^{n-2}$ = $\frac{2^{n}}{4}$ For x = 3, f(x) = $\binom{3}{2}*2^{n-3}$ = $\frac{3}{8}*2^{n}$ For x = 4, f(x) = $\binom{4}{2}*2^{n-4}$ = $\frac{3}{8}*2^{n}$ For x = 5, f(x) = $\binom{5}{2}*2^{n-5}$ = $\frac{5}{16}*2^{n}$ $\frac{5}{16} < \frac{3}{8}$ after that value of f(x) won't increase. So max value of f(x) = $\frac{3}{8}*2^{n}$ 1 votes 1 votes Please log in or register to add a comment.