Question

. Solve the following linear programming problem graphically :
Maximise Z = 7.5x + 5y
Subject to 2x + y ≤ 60
2x + 3y ≤ 120
x ≤ 20, x, y ≥ 0

Answer 1

❥︎Solution :-

Maximise Z = 7.5x + 5y

Subject to 2x + y ≤ 60

2x + 3y ≤ 120

x ≤ 20, x, y ≥ 0

❥︎Step :- 1

Convert these inequalities to equations, we get

2x + y = 60.......(1)

2x + 3y = 120.....(2)

x = 20 ......(3)

❥︎From line (1), point lies on the line are

$\begin{gathered}\boxed{\begin{array}{c|c|c}\bf x&\sf 0&\sf 30&\\{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}\\\bf y&\sf 60&\sf 0&\end{array}}\end{gathered}$

❥︎From line (2), points lies on the line are

$\begin{gathered}\boxed{\begin{array}{c|c|c}\bf x&\sf 0&\sf 60&\\{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}\\\bf y&\sf 40&\sf 0&\end{array}}\end{gathered}$

Plot the lines on graph paper, we get the shaded region as feasible region.

❥︎ So, corner points of the feasible region are

0 (0, 0)

A (20, 0)

B (20, 20)

C (15, 30)

D (0, 40)

❥︎ To find the maximum value of Z, we need to find the value of Z at the corner points.

[tex]

\begin{gathered}\boxed{\begin{array}{ccccc}\sf Corner \: point&\sf value \: of \: Z&\sf\sf &\sf \\\frac{\quad \qquad\qquad}{}&\frac{\quad \qquad \qquad}{}&\\\sf 0 (0, 0)&\sf 0&\sf &\sf &\sf \\\\\sf A (20, 0) &\sf 150&\sf &\sf &\sf \\\\\sf B (20, 20) &\sf 250&\sf &\sf &\sf \\\\\sf C (15, 30)&\sf 212.5&\sf &\sf &\sf \\\\\sf D (0, 40)&\sf 200 &\sf &\sf &\sf \\\frac{\qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad}{}&\frac{\qquad\qquad\qquad}{}&\frac{\qquad \qquad\qquad}{}&\frac{\qquad \qquad\qquad}{}\\\sf \sf \ \sf \end{array}}\end{gathered}}

❥︎ So, it means

\bf \:Maximum \: Value \: of \: Z = 250 \: at \: B(20, 20).