Applied exercises: Linear functions

For an increase to 140,000 copies, determine by how much the circulation has changed:

$\mathrm{\Delta }N=140000-120000=20000$

Determine the price change per additional copy $a$ from the problem definition: For 5000 additional copies, the price change is -0.2 €, so

$a=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }N}=\frac{-0.2}{5000}=-0.00004$

Use this rate to determine the price change induced by $\mathrm{\Delta }N$ copies.

$\mathrm{\Delta }p=a\cdot \mathrm{\Delta }N=-0.00004\cdot 20000=-0.8$

Use this to calculate the new price.

$p_{\mathrm{new}}=p_{\mathrm{old}}+\mathrm{\Delta }p=2.2-0.8=1.4$

So, with a circulation of 140,000 copies, the price is 1.40 € per copy.

Suppose, the price has fallen to 1.50 €. The price change then is:

$\mathrm{\Delta }p=p_{\mathrm{new}}-p_{\mathrm{old}}=1.5-2.2=-0.7$

Determine the number $b$ of additional copies per increase of the price by 1 €. (This is the inverse of the number $a$ in the first exercise part.)

$a=\frac{\mathrm{\Delta }N}{\mathrm{\Delta }p}=\frac{-5000}{0.2}=-25000$

Use this rate to determine the change of the circulation.

$\mathrm{\Delta }N=a\cdot \mathrm{\Delta }p=(-25000)\cdot (-0.7)=17500$

Determine the new circulation.

$N_{neu}=N_{alt}+\mathrm{\Delta }N=120000+17500=137500$

So with a price of 1.50 € per copy, the circulation will be at 137,500 copies.