The power ($P$) of an engine is measured in horsepower using the formula: $$P = \frac{R \times T}{5252}$$ where $R$ is the engine speed measured in revolutions per minute (RPM) and $T$ is the torque measured in appropriate units - Leaving Cert Mathematics - Question 8 - 2019

Starting from the original formula:

$P = \frac{R \times T}{5252}$

Multiply both sides by 5252:

$5252P = R \times T$

Now, divide both sides by $T$:

$R = \frac{5252P}{T}$

Given the initial loss of €4000 and reducing the loss by €250 each month, the profit can be expressed as:

$T_n = -4000 + (n-1)250$

Simplifying this results in:

$T_n = 250n - 4250$

To find the profit in month 25, substitute $n = 25$ into the profit formula:

$T_{25} = 250(25) - 4250$

Calculating this gives:

$T_{25} = 6250 - 4250 = 2000 €$

To find when the company breaks even, set the profit formula to zero:

$250n - 4250 = 0$

Solving for $n$ gives:

$250n = 4250$

Thus,

$n = \frac{4250}{250} = 17$

The company breaks even in month 17.

The total profit $S_n$ after $n$ months can be calculated as:

$S_n = \frac{n}{2}[-4000 + (n - 1)250]$

This simplifies to:

$S_n = \frac{n}{2}[-8000 + 250n]$

To find the total profit at the end of month 37:

$S_{37} = \frac{37}{2}[-8000 + 250 \times 37]$

Calculating inside the brackets first:

$S_{37} = \frac{37}{2}[-8000 + 9250] = \frac{37}{2}[1250]$

Thus,

$S_{37} = 37 \times 625 = 23125 €$