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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. $\left(\dfrac{4}{3} , \dfrac{20}{3}\right)$
edited | 424 views

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

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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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

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

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

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