JEE PYQ: Linear Programming Question 1
Question 1 - 2021 (17 Mar Shift 1)
The maximum value of $z$ in the following equation $z = 6xy + y^2$, where $3x + 4y \le 100$ and $4x + 3y \le 75$ for $x \ge 0$ and $y \ge 0$ is ______ (Round off to the Nearest Integer)
Type: Numerical
Show Answer
Answer: 904
Solution
$z = 6xy + y^2 = y(6x + y)$
$3x + 4y \le 100$
$4x + 3y \le 75$
$x \le (i)$
$Z \le \frac{1}{4}(225y - 7y^2) \le \frac{(225)^2}{2 \times 4 \times 7}$
$= \frac{50625}{56}$
$\approx 904.0178$
$\approx 904.02$
It will be attained at $y = \frac{225}{14}$
BETA
