Logical bulk pricing rate decrease

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Let's say I have items that I can sell up to 40 units at a time. If the sale is the maximum of 40, the price is $600 which comes comes out to $15 per item. I would like the price to be significantly higher if only 1-2 items are bought. How would I figure out:

  • 1 = $?

  • 2 = $?

  • 3 = $?

  • ...

  • 40 = $600

and to have the prices proportionally get lower each time. What would I need to start with for my 1st item (in case the customer only wanted one).

This data would eventually turn into slider widget on a website like one here where you can see what I consider my less than adequate attempt at this.

Thanks!

2

There are 2 best solutions below

1
On

To me, it seems like you are wanting something like this:

Start off by tacking on an additional $100\%$ of the item cost to the first item and keep decreasing it by a constant rate until $0\%$ is added to $40$ items. That is mathematically, consider a function $f:\{1,2,3,...,40\}\subset\mathbb N\to \mathbb R$ defined by: $$f(n)=15n+\left[15n\left(\frac{39-(n-1)}{39}\right)\right]$$ where $n$ is the number of items.
This will give you that one item will be $\$15 + \$15=\$30$. The rate thereafter will be decreasing by $\frac{1}{39}\approx.0256$ for each item added. And for the max sale of $40$ will not be charged anything extra only the price of $40\ \text{items}*\$15=\$600.$ For example, for 9 items, the total sale price will be $15(9)+$ roughly $79.49\%$*$15(9)=\$242.31.$ Let me know if you need clarification or if I did not understand your question completely.

0
On

You can also just use:

A = price of the first item

B = price of items if you buy 40

then the formula becomes $$f(n)=B \times n+(A-B) \times\left[ n \times \left(\frac{40 - n}{39}\right)\right]$$ where $n$ is the number of items.

PS this is just a simplification of jnh's answer