[soyuz02]

Hi,

I am working on an assignment for University and I don't seem to be getting anywhere.

The question states that a company charges 200$ for each order of 150 or fewer shares. The cost to the buyer on every share is reduced by 1$ for each order in excess of 150.

For what size order is the revenue maximum?

Ive come up with the following but it seemed to be too easy:

Let the number of shares = s

I have deduced that the company charges 200/150 so 4/3 per share.

Total Revenue=200 + 4/3s - s

The 200 is the min charge which the company seems to apply, 4/3s is what the company charges per share and -s subtracts 1 x number of shares from revenue.

Any idea if this is correct?
Where would I go from here?

Normally I would have expected to come up with a quadratic equation which I could then have solved and differentiated to find the values.

Thank you very much in advance!

[galactus]

You know Revenue is the number sold times the cost per item.





Set to 0 and solve for x, we get x=25.

Now, you can see the number to sell to maximize revenue?

[soyuz02]

You're a mircale!

Id like to learn out of this however so maybe you can just quickly explain why or how you got the revenue equation in the first place, (200-x)(150+x).

Thank you v much!

[galactus]

Revenue = cost per item times items sold.

Let x = the number of units added. For every x items added to 150, the price drops by
$1x dollars.

Hence,

Using what we found from differentiating, we find $175 should be charged per order and sell 175 of them.

175*175=$30625 total max revenue.


Does this help? I don't know how to explain it any better.


Cheers,
Cody

[soyuz02]

Thanks, you've helped me a great deal :)

[galactus]

You're very welcome. A large part of the ingrates who post never have the common courtesy to say thanks.