Question

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given in below Image. Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of typeA and 30 of type B should be manufactured in a day to get maximum profit.

Answer 1

Solution:

Let x toys of type A and y toys of type B be manufactured.

So

$x \geqslant 0 \: \: \: \: eq1 \\ \\ y \geqslant 0 \: \: \: \: eq2 \\ \\ 12x + 6y \leqslant 360 \\ \\ 2x + y \leqslant 60 \: \: \: \: \: eq3 \\ \\ 18x + 0y \leqslant 360 \\ \\ x \leqslant 20 \: \: \: \: eq4\\ \\ 6x + 9y \leqslant 360 \\ \\ 2x + 3y \leqslant 120 \: \: \: eq5$
profit function

$Z = 7.5x + 5y$
on plotting all eqs1 to 5 ,we get total five points in common to all region

A(20,0)

B(0,0)

C(20,20)

D(15,30)

F(0,40)

Put all these points in the profit function and check at which point it will be maximum

$at \: A = Z = 7.5(20) + 5(0) = 150 \\at \: B= Z= 7.5(0) + 5(0) = 0 \\at \: C = Z = 7.5(20) + 5(20) = 250 \\at \: D= Z = 7.5(15) + 5(30) = 262.5 \\at \: F = Z = 7.5(0) + 5(40) = 200$

Hence profit is maximum at x= 15,y = 30

Proved

Answer 2

This is the detailed solution of your question.