Explanation

For a rectangular channel to have maximum discharge for a given cross-sectional area, the hydraulic radius must be maximized. This occurs when the width is twice the depth (or depth is half the width) only if we are optimizing for hydraulic radius, but for maximum discharge with a fixed area, the condition is that the wetted perimeter is minimized. Minimizing the wetted perimeter for a rectangle leads to a semi-square section where the width is twice the depth. Wait, let's re-verify. Condition for most economical rectangular channel: B = 2D (Width = 2 * Depth). This means Depth = Width / 2. So Depth is half of its width. Let's check the options again. A: Twice the width -> D = 2B? No. B: Equal to its width -> D = B? No. C: Half of its width -> D = B/2? Yes, this matches B=2D. D: one-third of its with -> D = B/3? No. So the current answer C is correct. The depth should be half the width.