(a) \( f(x)=2400=2 x \) (b) \( A(x)=2400 x-2 x^{2} \) (d) Finst write the expression for the derivetive used to find the \( x \) value that maximizes area \( \frac{d A}{d x}=-4 x+2400 \) The \( x \)-value leading to the maximum area is \( \square \square \nabla \)

To find the \( x \)-value that maximizes the area, we need to set the derivative equal to zero. So we solve \( -4x + 2400 = 0 \). This gives us \( x = 600 \), which means when \( x \) is 600, the area is maximized! Now, let's visualize the situation. Suppose this relates to a rectangular garden where the area is expressed in terms of its length and width. By optimizing one dimension, you're essentially maximizing your gardening space, making your plot not just more beautiful, but also more productive for growing those favorite veggies! What a win-win, right?