This is an optimization problem where Omar wants to maximize his revenue based on the price of soda cans and customer demand. Here's a breakdown of the problem:

Given Information:

• Current price per can = $0.65
• Current number of cans sold per week = 600 cans
• Decrease of 100 customers per $0.05 price increase
• Increase of 100 customers per $0.05 price decrease

Let:

• $p$ = price per can (in dollars)
• $q(p)$ = quantity of soda cans sold at price $p$

Relationship between price and quantity:

• For every $0.05 increase in price, he loses 100 customers (so the quantity decreases).
• For every $0.05 decrease in price, he gains 100 customers (so the quantity increases).

We can express the relationship as: $q(p) = 600 + 200 \cdot (0.65 - p)$

The term $(0.65 - p)$ represents the change in price relative to the original price. The multiplier 200 represents the change in customers (100 per $0.05, and $0.05 = 1 unit change).

Revenue function:

Omar's revenue $R(p)$ is the product of the price and quantity sold: $R(p) = p \cdot q(p) = p \cdot \left(600 + 200 \cdot (0.65 - p)\right)$

This is the revenue function we can use to determine the optimal price to maximize revenue.

Would you like me to continue and find the optimal price Omar should charge to maximize his revenue?

Here are 5 relative questions to expand on this:

1. What is the effect of a $0.10 price decrease on the quantity sold?
2. What would be Omar's revenue if he increased the price by $0.10?
3. How would customer behavior impact the vending machine's revenue over time?
4. How does maximizing revenue differ from maximizing profit?
5. What would happen if there were no change in the number of customers for price increases?

Tip: In optimization problems, revenue is often maximized at the point where price balances with demand.