A manufacturer finds that the revenue generated by selling \( x \) units of a certain commodity is given by the function \( R(x)=100 x-0.2 x^{2} \), where the revenue \( R(x) \) is measured in dollars. What is the maximum revenue, and how many units should be manufactured to obtain this maximum? \( \$ \square \) units Need Help? Watch it Master lt

To find the maximum revenue, we need to determine the vertex of the quadratic function \( R(x) = 100x - 0.2x^2 \). The vertex can be found using the formula \( x = -\frac{b}{2a} \) for the function \( ax^2 + bx + c \). Here, \( a = -0.2 \) and \( b = 100 \). So, \( x = -\frac{100}{2 \times -0.2} = 250 \). Plugging this back into the revenue function, we calculate \( R(250) = 100(250) - 0.2(250^2) = 25000 - 12500 = \$12500 \). Thus, the maximum revenue is \$12,500 when 250 units are manufactured. Think of maximizing revenue as an art form; you want to hit the sweet spot for production. Too few units and you're missing out; too many, and you're sinking resources into unsold stock! Finding that balance not only enhances profits but sets the stage for future growth, allowing more room for innovation and market expansion. Remember, in business, it's all about that perfect tune—the right product at the right quantity!