The maximum value of $z$ in the following equation $z=6 x y+y^{2}$, where $3 x+4 y \leq 100$ and $4

Question

The maximum value of $z$ in the following equation $z=6 x y+y^{2}$, where $3 x+4 y \leq 100$ and $4 x+3 y \leq 75$ for $x \geq 0$ and $y \geq 0$ is

Professor PC Thomas Classes · Answer

(904): Feasible region is shaded in the graph.\ Let the maximum value of $6 x y+y^{2}=M$\ For a solution in feasible region,\ $6 x y+y^{2}=M$ and $4 x+3 y=75$ must have at least 1 possible solution.\  $	herefore \quad y^{2}+6 y\left(\frac{75-3 y}{4}ight)-M=0$\ $\Rightarrow \quad \frac{7}{2} y^{2}-\frac{225}{2} y+M=0$\ Now, discriminant $\geq 0$\ $\Rightarrow\left(\frac{225}{2}ight)^{2} \geq 4 	imes \frac{7}{2} 	imes M$\ $\Rightarrow \quad M \leq \frac{(225)^{2}}{4 	imes 2 	imes 7} \cong 904$\