QUESTION 3
D1 JANUARY 2012 EXAM
PAPER
Federico Midolo
In order to do this, we will
need to sketch the
constraints, represented by
the inequailites, on a graph.
Let’s start by treating them
as normal equations, to find
their x and y intercepts.
Note: the sign has
changed to equal, so
that it is easier to find
the x and y intercepts
x-intercept
y-intercept
x+y=11
when x=0, y=11;
when y=0, x=11
(11, 0)
(0, 11)
3x+5y=39
when
x=0,
y=39÷5=7.8;
Now we
can
plot these lines
when
y=0, x=39÷3=13
on a graph(next
slide).
(13, 0)
(0, 7.8)
x+6y=39
when x=0, y=39÷6=6.5;
when y=0, x=39
(39, 0)
(0, 6.5)
y
Note: always label
the plotted lines.
x-intercept
20
Shade out the
unaccaptable
region, to keep
the feasible region
clear and easy to
identify.
15
y-intercept
We now need(11,
to 0)
plot the (0,
objective
11)
x+y=11
line 2x+3y and virtually move it across
our
graph to find
the intersection
(0, 7.8)
(13, 0)
3x+5y=39
point located furthest away (as we
are
asked to maximise).
(0, 6.5)
(39, 0)
x+6y=39
2x+3y (x and y intercepts)
When x=0, y=coefficient of x=2 (0,2)
When y=0, x=coefficient of y=3 (3,0)
x+y=11
10
3x+5y=39
x+6y=39
5
0
This is the maximum point of 2x+3y,
when subject to these constraints. We
can find its coordinates accurately, by
solving the simultaneous equation
x+y=11 and 3x+5y=39 (next page).
10
20
2x+3y
30
40
x
x+y=11 and 3x+5y=39
x+y=11
y=11-x
3x+5(11-x)=39
3x+55-5x=39
2x=16
x=8
y=11-8=3
The answer is
x=8, y=3