A company makes and sells fibre optic cable - Leaving Cert Mathematics - Question 8 - 2017

Set the profit equal to €8350:

$P(x) = 275x - x^2 - 2000 = 8350.$
Rearranging gives:

$275x - x^2 = 10350$

or:

$-x^2 + 275x - 10350 = 0.$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
with $a = -1$, $b = 275$, and $c = -10350$, we find:

$x = \frac{-275 \pm \sqrt{(275)^2 - 4(-1)(-10350)}}{2(-1)}.$
Calculating this results in:

$x = \frac{275 \pm 145}{2} = \{210, 45\}.$
Since the production limit is 200 km, the valid solution is $x = 45$ km.

Using the profit function to calculate:

• For $x = 50$: $P(50) = 275(50) - (50)^2 - 2000 = 9250$
• For $x = 60$: $P(60) = 275(60) - (60)^2 - 2000 = 10350$
• For $x = 70$: $P(70) = 275(70) - (70)^2 - 2000 = 12350$
• For $x = 80$: $P(80) = 275(80) - (80)^2 - 2000 = 14550$
• For $x = 90$: $P(90) = 275(90) - (90)^2 - 2000 = 16800$
• For $x = 100$: $P(100) = 275(100) - (100)^2 - 2000 = 19000$
  Thus, the table is completed as:

Number of km of cable sold $(x)$	50	60	70	80	90	100
Profit (€)	9250	10350	12350	14550	16800	19000

Using the completed table, plot the points on a graph where:

• $x = 50, y = 9250$
• $x = 60, y = 10350$
• $x = 70, y = 12350$
• $x = 80, y = 14550$
• $x = 90, y = 16800$
• $x = 100, y = 19000$. Connect these points to visualize the profit function.

From the graph, locate where the profit is €10,000 and €14,000. It appears between:

• Lower bound: approximately $55$ km
• Upper bound: approximately $83$ km.

We need to differentiate the profit function:

$\frac{dP}{dx} = 275 - 2x = 105.$
Solving for $x$ gives:

$2x = 275 - 105$
$2x = 170$

thus, $x = \frac{170}{2} = 85$ km.
The company sells 85 km of cable when the profit is increasing at €105 per km.