A small factory makes bars of soap - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 2

Given the profits, we can derive two equations:

1. For 800 bars:

$y = 2(800) - 500$

which simplifies to:

$y = 1600 - 500$

leading to:

$y = 1100$

2. For 300 bars:

$y = 2(300) + 80$

which simplifies to:

$y = 600 + 80$

leading to:

$y = 680$

Now, substituting into the linear equation $y = k + cx$ gives us:

$1100 = k + 800c$ $680 = k + 300c$

Solving these two equations by elimination leads to:

After subtracting the second from the first,

$1100 - 680 = 800c - 300c$ $420 = 500c$ $c = 0.84$

By substituting $c = 0.84$ back into either equation, say $680 = k + 300(0.84)$,

we find:

$680 = k + 252$ $k = 680 - 252 = 428$

Thus:

$y = 0.84x + 428$

The value 0.84 represents the variable cost of producing each additional bar of soap. It quantifies the incremental cost incurred for each unit produced, signifying that for each bar made, the factory incurs an expense of £0.84.

To find the least number of bars, we set the profit to be greater than 0:

$ext{Revenue} - ext{Cost} > 0$

The revenue from selling $x$ bars is:

$2x$

The total cost is:

$y = 0.84x + 428$

Setting the equation for profit gives:

$2x - (0.84x + 428) > 0$

Simplifying:

$2x - 0.84x - 428 > 0$ $1.16x > 428$ x > rac{428}{1.16}

Calculating:

$x > 369.0$

Thus, the least number of bars that must be produced is 370.