Question

Suppose a manufacturer of printed circuits has a stock of $200$ resistors, $120$ transistors and $150$ capacitors and is required to produce two types of circuits.
Type A requires $20$ resistors, $10$ transistors and $10$ capacitors
Type B requires $10$ resistors, $20$ transistors and $30$ capacitors.
If the profit on type $A$ circuits is $E5$ and that on type $B$ circuits is $E12$, how many of each circuit should be produced in order to maximize profit?

Answer 1

Given:

	Resistors	Transistors	Capacitors
Type A$\left(x\right)$	20	10	10
Type B$\left(y\right)$	10	20	30
Total	200	120	150

Maximize $Z=5x+12y$

Solution:

Let type A be $x$ and let type be $y$

$i)20x+10y\le 200ii)10x+20y\le 120iii)10x+30y\le 150$ constraints

i)$20x+10y=200$

$x$	0	10
$y$	20	0

Point A(0,20), B(10,0)
ii)$10x+20y=120$

$x$	0	12
$y$	6	0

Point C(0,6) D(12,0)
iii)$10x+30y=150$

$x$	0	15
$y$	5	0

Point E(0,5) F(15,0)

iv)

$\frac{\frac{20x+10y=20010x+20y=120}{400+200=4000\pm 100x\pm 200=\pm 1200}}{300=2800x=9.3}20x+10y=20020\left(9.3\right)+10y=20010y=14y=1.4$ G(9.3,1.4)

$\frac{10x+20y=120\pm 10x\pm 30y=\pm 150}{-10y=-30y=3}10x+20y=12010x+60=12010x=60x=6$ Point H(6,3)

The feasible region $CGBO$

Points	$z=5x+12y$
O(0,0)	0
C(0,6)	72 $\to$ Maximum
G(9.3,1.4)	63.3
B(10,0)	50

The maximum value is 72 and the points $(x,y)$ are

(0,6)$\Rightarrow$ $x=$ $y=6$