A dog food producer reduced the price of a dog food. With the price at $8 the average monthly sales has been 22000 . When the price dropped to $7, the average monthly sales rose to 25000 . Assume that monthly sales is linearly related to the price. What price would maximize revenue?

Step 1 ：We are given that the price of dog food was initially $8 and the average monthly sales were 22000. When the price dropped to $7, the average monthly sales rose to 25000. We are assuming that the monthly sales is linearly related to the price.

Step 2 ：We can use these two points to find the equation of the line that represents the relationship between price (p) and quantity sold (q). The slope (m) of the line can be calculated as \(m = \frac{q_2 - q_1}{p_2 - p_1} = \frac{25000 - 22000}{7 - 8} = -3000.0\)

Step 3 ：The y-intercept (b) of the line can be calculated by substituting one of the points into the equation \(q = mp + b\). Using the point (8, 22000), we get \(b = q - mp = 22000 - (-3000.0 * 8) = 46000.0\)

Step 4 ：So, the equation of the line is \(q = -3000.0p + 46000.0\)

Step 5 ：The revenue (R) is calculated as the product of the price and the quantity sold. So, \(R = p * q = p * (-3000.0p + 46000.0)\)

Step 6 ：To find the price that maximizes revenue, we take the derivative of the revenue function with respect to price and set it equal to zero. The derivative of the revenue function is \(R' = -6000.0p + 46000.0\)

Step 7 ：Solving the equation \(-6000.0p + 46000.0 = 0\) for p, we get \(p = \frac{46000.0}{6000.0} = 7.66666666666667\)

Step 8 ：Final Answer: The price that would maximize revenue is approximately \(\boxed{7.67}\)