Solve the following linear programming problem graphically:
Maximize $Z = 7 x + 10 y$
subject to the constraints
$4 x + 6 y \leq 240$
$6 x + 3 y \leq 240$
$x \geq 10$
$x \geq 0, y \geq 0$

Open in App

Solution

The LPP given is:
Maximize $7 x + 10 y$ subject to contsraints
$4 x + 6 y \leq 240 \Rightarrow 2 x + 3 y \leq 120$
$6 x + 3 y \leq 240 \Rightarrow 2 x + y \leq 80$
$x \geq 10$
$y \geq 0$

Plot the graphs of $2 x + 3 y = 120, 2 x + y = 80, x = 10$ . The shaded portion is the feasible region of the solution.
The corner points of the feasible region are $(10, 0), (40, 0), (30, 20)$ and $(10, \frac{100}{3})$
Examining the objective function at these values:

$(x, y)$ Value of objective function: $7 x + 10 y$
$(10, 0)$ $70$
$(40, 0)$ $280$
$(30, 20)$ $410$
$(10, \frac{100}{3})$ $403.33$
From the values, we can see that $7 x + 10 y$ is maximum when $x = 30, y = 20$ . The maximum value is $410$ .

Suggest Corrections

Similar questions

Z = 7 x + 10 y

4 x + 6 y \leq 240 6 x + 3 y \leq 240 x \geq 10 x \geq 0, y \geq 0

x + y \leq 50 3 x + y \leq 90 x, y \geq 0

Z = 60 x + 40 y

x + 2 y \leq 12

2 x + y \leq 12

x + \frac{5}{4} y \geq 5; x \geq 0, y \geq 0

z =

x +

y

x +

y \geq

x +

y \geq

x +

y \leq

x \geq

y \geq

Z = 3 x + 4 y

x + y \leq 4, x \geq 0, y \geq 0

Join BYJU'S Learning Program

$(x, y)$	Value of objective function: $7 x + 10 y$
$(10, 0)$	$70$
$(40, 0)$	$280$
$(30, 20)$	$410$
$(10, \frac{100}{3})$	$403.33$