QJEEM21Ma89c7: JEE Main 2021 Mathematics - Class 11 JEEM21Ma89c7 Solution: The maximum value of z in the following equation z = 6xy + y 2…

3x + 4y $\le$ 100

4x + 3y $\le$ 75

x $\ge$ 0, y $\ge$ 0

Feasible region is shown in the graph

Let maximum value of 6xy + y² = c

For a solution with feasible region,

6xy + y² = c and 4x + 3y = 75 must have at least one positive solution.

${y^2} + 6y\left( {{{75 - 3y} \over 4}} \right) - c = 0$

$\Rightarrow {7 \over 2}{y^2} - {{225} \over 2}y + c = 0$

$\Rightarrow {\left( {{{225} \over 2}} \right)^2} \ge 4.{7 \over 2}.c$

$\Rightarrow c \le {{{{225}^2}} \over {56}} \approx 904$