HighestRevenue | Topcoder Archive

Problem Statement

The Demand for a product is the relationship between the price we charge and the quantity we can sell at that price. Our problem is to get the most possible revenue (price times quantity) by choosing the price wisely.

The Demand is given as p = a - b * q where p is the maximum price we can charge if we want to sell q items, and a and b are constants. q is number of items, and p is cents per item. Both must be non-negative integers.

Create a class HighestRevenue that contains a method optimalPrice that takes a and b as inputs and returns the price to charge that will result in the maximum revenue (p*q) possible. If two different prices result in the maximum revenue, return the larger price.

Definition

Class:: HighestRevenue
Method:: optimalPrice
Parameters:: int, int
Returns:: int
Method signature:: int optimalPrice(int a, int b)
(be sure your method is public)

Notes

The maximum revenue may be 0 (in which case your method should return a)
Both p and q must be non-negative integers

Constraints

a and b are between 1 and 10,000 inclusive

Examples

```
7000
```
```
50
```
```
Returns: 3500
```
p = 7000 - 50q. At a price of 3500, the quantity demanded will be 70 and the total revenue will be 70 * 3500 = 245,000. The next higher price we could charge would be 3550, resulting in the sale of 69 items generating a revenue of 69 * 3550 = 244,950. The next lower price we could charge would be 3450 generating sales of 71 items and revenue of 71 * 3450 = 244,950.
```
70
```
```
35
```
```
Returns: 35
```
p = 70 - 35q The only prices that correspond to valid values for q are 70, 35, and 0. Only a price of 35 yields a positive revenue.
```
70
```
```
80
```
```
Returns: 70
```
p = 70 - 80q At a price of 70, 0 units will be "sold" yielding a revenue of 0. At no other price will the corresponding quantity be a non-negative integer.
```
5625
```
```
125
```
```
Returns: 2875
```
Both p=2875 and p=2750 yield the same maximal revenue of 63250, so the larger one is returned.

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