Linear Programming — JEE Main · 2021 · 17 Mar 2021 · Shift 1 Question with Solution

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 ____

Answer: 904

Solution

The objective

z = 6 x y + y^{2}

increases with both

x

and

y

, so its largest value over the feasible region lies on the boundary. The lines

3 x + 4 y = 100

and

4 x + 3 y = 75

meet where

x = 0, y = 25

, and at that corner

z = y^{2} = 625

. Along the edge

4 x + 3 y = 75

one may write

x = \frac{75 - 3 y}{4}

, so that

z = 6 y \cdot \frac{75 - 3 y}{4} + y^{2} = 112.5 y - 3.5 y^{2}

. Its derivative

112.5 - 7 y

vanishes at

y = \frac{112.5}{7}

, where

x = \frac{75 - 3 y}{4} > 0

keeps the point feasible, and substituting back gives

z = \frac{112. 5 ^{2}}{14} \approx 904

. The maximum value is

904

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer