13 votes 13 votes $X$ and $Y$ are two positive real numbers such that $2X+Y \leq 6$ and $X + 2Y \leq 8.$For which of the following values of $(X,Y)$ the function $f(X,Y)=3X + 6Y$ will give maximum value ? $\left(\dfrac{4}{3} , \dfrac{10}{3}\right)$ $\left(\dfrac{8}{3} , \dfrac{20}{3}\right)$ $\left(\dfrac{8}{3} , \dfrac{10}{3}\right)$ $\left(\dfrac{4}{3} , \dfrac{20}{3}\right)$ Quantitative Aptitude gate2013-ce quantitative-aptitude maxima-minima + – Akash Kanase asked Feb 16, 2016 • edited Jun 8, 2018 by Milicevic3306 Akash Kanase 3.0k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 15 votes 15 votes $f(x,y)=3x+6y=3(x+2y)$ maximum value of $x+2y$ is given by equation $2$ which is $8$ $(x+2y<=8)$ i.e maximum value of $f(x,y)=3*(8)=24$ $f(x,y)=24$ can be obtained by putting $x=4/3$ & $y=10/3$ So, answer is $(A)$. srivivek95 answered Apr 9, 2016 • edited Jun 8, 2018 by Milicevic3306 srivivek95 comment Share Follow See all 2 Comments See all 2 2 Comments reply MiNiPanda commented Nov 20, 2018 reply Follow Share Solving without options: f(x,y)=3X+6Y max of f(x,y) will be obtained when both X and Y are having their max values. 2X+Y≤6 --> (1) max(2X+Y)=6 X+2Y≤8 -->(2) max(X+2Y)=8 Adding (1) and (2), 3X+3Y≤14 => X+Y≤14/3 => max(X+Y)=14/3 ---> (3) Max of f(X,Y)= max(3X+6Y)= max (3(X+2Y)) = 3*max(X+2Y) =3*8=24 ---> (4) Max of f(X,Y)= max(3X+6Y)=max(3(X+Y) + 3Y) = 3*max(X+Y) + 3*max(Y) = 3*14/3 + 3*max(Y) --->(5) Equating (4) and (5), 24=14+3*max(Y) => max(Y)=10/3 Placing Y in Eq (3), max(X)= 14/3-10/3=4/3 5 votes 5 votes VIDYADHAR SHELKE 1 commented Jan 29, 2020 reply Follow Share we can also solve through linear programing method 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes I think ans is A as only A satisfy both conditions(i.e 2x+y<=6 and x+2y<=8). richa07 answered Feb 26, 2016 richa07 comment Share Follow See all 0 reply Please log in or register to add a comment.