Question

Solve the following LPP graphically:

Maximise Z= 2x + 3y, subject to x + y ≤ 4, x ≥ 0, y ≥ 0

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Answer 1

$\large\underline{\sf{Solution-}}$

Given objective function is

Maximise Z = 2x + 3y

Constraints are x + y ≤ 4, x ≥ 0, y ≥ 0

Let we first plot the line

$\rm \: x + y = 4$

Substituting 'x = 0' in the given equation, we get

$\rm \: 0 + y = 4$

$\rm\implies \:\rm \: y = 4$

Substituting 'y = 0' in the given equation, we get

$\rm \: x + 0 = 4$

$\rm\implies \:x = 4$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}$

Now, See the graph in attachment.

OAB is the required feasible region and the corner points of the feasible region are A (4, 0), B (0, 4) and O (0, 0).

Now, The value of Z at corner points are as follow :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: point & \bf Z = 2x + 3y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf A(4,0) & \sf 8 + 0 = 8 \\ \\ \sf B(0,4) & \sf 0 + 12 = 12 \\ \\ \sf O(0,0) & \sf 0 \end{array}} \\ \end{gathered}$

This implies Maximum value of Z is 12 at B (0, 4)

Answer 2

Answer:

if it's u in dp then i guess u are cute and bubbly