a) Let's represent the price of a garden ornament as "p" and the number of ornaments sold as "n".
Since the price of a garden ornament decreases by $0.50 each time, we can express the price as: p = 4 - 0.50x, where "x" is the number of $0.50 decreases.
Similarly, the number of ornaments sold can be expressed as: n = 120 + 20x, where "x" is the number of $0.50 decreases.
b) The revenue is calculated by multiplying the price of a garden ornament by the number of ornaments sold. Therefore, the revenue equation can be written as: R = p * n.
Substituting the expressions from part a) into the revenue equation, we get: R = (4 - 0.50x) * (120 + 20x).
c) To find the price the artisan should charge to maximize revenue, we need to find the value of "x" that maximizes the revenue equation. We can achieve this by finding the maximum of the revenue equation, which is a quadratic equation.
First, let's expand and simplify the revenue equation: R = (4 - 0.50x) * (120 + 20x) = 480 + 80x - 60x - 10x^2.
Rearranging the equation to write it in standard quadratic form: R = -10x^2 + 20x + 480.
To find the maximum revenue, we need to find the vertex of the quadratic equation. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where "a" is the coefficient of the x^2 term and "b" is the coefficient of the x term.
In our equation, a = -10 and b = 20, so the x-coordinate of the vertex is x = -20 / (2(-10)) = 1.
Substituting the value of x = 1 back into the revenue equation, we can find the maximum revenue: R = -10(1)^2 + 20(1) + 480 = -10 + 20 + 480 = 490.
Therefore, to maximize revenue, the artisan should charge a price that corresponds to x = 1, which means a decrease in price by $0.50. The original price of $4 minus $0.50 would give the price that maximizes revenue, which is $3.50 per garden ornament.