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$X$ 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 ?

  1. $\left(\dfrac{4}{3} , \dfrac{10}{3}\right)$
  2. $\left(\dfrac{8}{3} , \dfrac{20}{3}\right)$
  3. $\left(\dfrac{8}{3} , \dfrac{10}{3}\right)$
  4. $\left(\dfrac{4}{3} , \dfrac{20}{3}\right)$
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2 Answers

Best answer
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$f(x,y)=3x+6y=3(x+2y)$

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 $(A)$.
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I think ans is A as only A satisfy both conditions(i.e 2x+y<=6 and x+2y<=8).

Answer:

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