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

Set the profit function equal to €8350:

$275x - x^2 - 2000 = 8350$

Rearranging gives:

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

Using the quadratic formula, we can calculate:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -1$, $b = 275$, and $c = -10350$. Calculating the discriminant:

$b^2 - 4ac = 275^2 - 4(-1)(-10350) = 75625 - 41400 = 34225$

Continuing with the quadratic formula:

$x = \frac{-275 \pm \sqrt{34225}}{2(-1)}$ $\sqrt{34225} = 185$

Thus, the roots are:

$x = \frac{275 \pm 185}{2}$

Calculating, we find:

$x = \frac{460}{2} = 230 \quad \text{(not possible, as production limit is 200)}$ $x = \frac{90}{2} = 45$

Therefore, the company sold 45 km of cable.

Using the function:

• For $x = 50$: $P(50) = 275(50) - 50^2 - 2000 = 9250$

• For $x = 60$: $P(60) = 275(60) - 60^2 - 2000 = 10900$

• For $x = 70$: $P(70) = 275(70) - 70^2 - 2000 = 12600$

• For $x = 80$: $P(80) = 275(80) - 80^2 - 2000 = 14300$

• For $x = 90$: $P(90) = 275(90) - 90^2 - 2000 = 16100$

• For $x = 100$: $P(100) = 275(100) - 100^2 - 2000 = 17900$

The completed table is:

Number of km of cable sold (x)	Profit (€)
50	9250
60	10900
70	12600
80	14300
90	16100
100	17900

From the graph, you would locate the points where the profit intersects with €10,000 and €14,000.

• For €10,000, the lower range (approx.) is at $x = 55$ km and the upper range at $x = 83$ km.

• For €14,000, the ranges can be roughly estimated via similar graph observations and would be around $x = 70$ km to $x = 90$ km.

First, differentiate the profit function:

$\frac{dP}{dx} = 275 - 2x$

Set this equal to €105 (the rate of increase in profit):

$275 - 2x = 105$

Solving for $x$ gives:

$2x = 275 - 105 \\ 2x = 170 \\ x = 85$

Thus, the number of kilometres of cable sold when the profit is increasing at €105 per km is 85 km.