Question

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs.17.50 per package on nuts and Rs.7 per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operates his machines for at the most 12 hours a day? Form the linear programming problem and solve it graphically.

Answer 1

Let x package nuts and y package bolts are produced
Let z be the profit function, which we have to maximize.
Here z = 17.50x + 7y ... (i) is objective function.
And constraints are
x + 3y ≤12 ...(ii)
3x + y ≤12 ...(iii)
x≥ 0 ...(iv)
y≥0 ...(v)
On plotting graph of above constraints or inequalities (ii), (iii), (iv) and (v) we get shaded region as feasible region having corner points A, O, B and C.

For coordinate of ‘C’ two equations
x + 3y = 12 ...(vi)
3x + y = 12 ...(vii) are solved

Applying (vi) × 3 – (vii), we get
3x + 9y - 3x - y = 36 - 12
=> 8y = 24 => y = 3 and x = 3
Hence coordinate of C are (3, 3).
Now the value of z is evaluated at corner point as

Therefore maximum profit is Rs.73.5 when 3 package nuts and 3 package bolt are produced.