WAEC 2020 · Paper 2 · Q4

(a)
Shade the region, $P$ , in the $x$ – $y$ plane which satisfies simultaneously the inequalities:

$x + y \ge 5, \qquad 2y - x \ge 0, \qquad x + 5y \le 21$
Model answer
Draw the three boundary lines (all solid, since every inequality includes equality): $x + y = 5$ through $(5, 0)$ and $(0, 5)$ ; $2y = x$ through $(0, 0)$ and $(6, 3)$ ; $x + 5y = 21$ through $(1, 4)$ and $(6, 3)$ . The region $P$ is the triangle with corners $(1, 4)$ , $(3\frac13, 1\frac23)$ and $(6, 3)$ . Shade it (or shade the unwanted sides) and label it $P$ .
(b)(i)
Use the diagram in 4(a) to find, on the region $P$ , the minimum value of $x$ ;
(b)(ii)
the maximum value of $(3x + 2y)$ .

Try it on a graph

The green region satisfies all three inequalities. Slide the objective line to find the maximum of 3x + 2y. Edit an inequality to see the region change.

Worked solution (try it first)

(a)

Draw the three boundary lines as solid lines: $x + y = 5$ through $(0, 5)$ and $(5, 0)$ .
$2y - x = 0$ , that is $y = \frac{x}{2}$ , through $(0, 0)$ and $(6, 3)$ .
And $x + 5y = 21$ through $(1, 4)$ and $(6, 3)$ .
Test a point, such as $(3, 3)$ : $3 + 3 = 6 \ge 5$ , $6 - 3 = 3 \ge 0$ and $3 + 15 = 18 \le 21$ , so it is in $P$ .
The region $P$ is the triangle containing $(3, 3)$ .
Shade it.
Its corners are where the lines meet: $x + y = 5$ and $x + 5y = 21$ give $4y = 16$ , so $(1, 4)$ .
$y = \frac{x}{2}$ and $x + y = 5$ give $\frac{3x}{2} = 5$ , so $\left(\frac{10}{3}, \frac53\right)$ .
$y = \frac{x}{2}$ and $x + 5y = 21$ give $\frac{7x}{2} = 21$ , so $(6, 3)$ .

(b)(i)

The point of $P$ furthest to the left is the corner $(1, 4)$ , so the minimum value of $x$ is $1$ .

(ii)

The largest value of $3x + 2y$ on the region is at a corner.
At $(1, 4)$ it is $11$ .
At $\left(\frac{10}{3}, \frac53\right)$ it is $13\frac13$ .
At $(6, 3)$ it is $24$ .
So the maximum is $24$ , at $(6, 3)$ .

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