Until recently, hamburgers at the city sports arena cost $\$ 3.40$ each. The food concessionaire sold an average of 4,000 hamburgers on game night. When the price was raised to $\$ 4.00$, hamburger sales dropped off to an average of 2,500 per night.
(a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue.
(b) If the concessionaire had fixed costs of $\$ 1,500$ per night and the variable cost is $\$ 0.70$ per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit.
(a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue.
The hamburger price that will maximize the nightly hamburger revenue is $\$$
(Round to the nearest cent as needed.)
(b) If the concessionaire had fixed costs of $\$ 1,500$ per night and the variable cost is $\$ 0.70$ per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit.
The hamburger price that will maximize the nightly hamburger profit is $\$ \square$.

Step 1 ：We are given two points on the demand curve: (3.4, 4000) and (4, 2500). We can use these points to find the equation of the line, which will be in the form y = mx + b, where m is the slope and b is the y-intercept.

Step 2 ：The slope m can be found using the formula \((y2 - y1) / (x2 - x1)\), which gives us m = -2499.9999999999995.

Step 3 ：The y-intercept b can be found by substituting one of the points into the equation and solving for b, which gives us b = 12499.999999999998.

Step 4 ：So, the equation of the demand curve is \(y = 12500.0 - 2500.0x\).

Step 5 ：The revenue function is the price times the quantity, or \(R = pxq\), which gives us \(R = x*(12500.0 - 2500.0x)\).

Step 6 ：To find the price that maximizes revenue, we take the derivative of the revenue function and set it equal to zero, which gives us \(12500.0 - 5000.0x = 0\).

Step 7 ：Solving this equation gives us the price that maximizes revenue, which is \(\boxed{2.50}\).

Step 8 ：To find the price that maximizes profit, we need to subtract the cost of producing the hamburgers from the revenue. The cost function is given by the fixed costs plus the variable cost times the quantity, or \(C = 1500 + 0.7q\).

Step 9 ：We can substitute the demand function for q in the cost function to get the cost in terms of price, which gives us \(C = 10250.0 - 1750.0x\).

Step 10 ：Subtracting this from the revenue function gives us the profit function, which is \(P = x*(12500.0 - 2500.0x) + 1750.0x - 10250.0\).

Step 11 ：Taking the derivative of the profit function and setting it equal to zero gives us \(14250.0 - 5000.0x = 0\).

Step 12 ：Solving this equation gives us the price that maximizes profit, which is \(\boxed{2.85}\).