Mathematics · 2021
JEE Main · 17 March 2021, Shift 1 · Q81
The maximum value of z in the following equation z=6 x y+y^2, where 3 x+4 y ≤ 100 and 4 x+3 y ≤ 75 for x ≥ 0 and y ≥ 0 is \ \ \ \.
The maximum value of $\displaystyle z$ in the following equation $\displaystyle z=6 x y+y^2$, where $\displaystyle 3 x+4 y \leq 100$ and $\displaystyle 4 x+3 y \leq 75$ for $\displaystyle x \geqslant 0$ and $\displaystyle y \geqslant 0$ is \_\_\_\_.
Official answer
904
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