I'm fairly sure that the error of 0.05 cm is for the 8 cm width. You should use 0.05 cm to be your change in width, or dw, and use 8 cm to be your width, or w.
A child's toy is a flat disk shaped rectangle with a semi-circular end. The length of the rectangle is three times the width and the width is measured to be 8 cm with an error of 0.05 cm. Use differentials to find the percentage error in the area.
The shape they are referring too looks similar to this flipped clockwise 90 degrees. Ignore the dimensions on the image it's from a quick google image search.
The bold/italic part is what confuses me. Is the total area 0.05 cm off? Or just the 8 cm of the vertical portion only? I'm leaning towards the 8 cm vertical portion only because they didn't specify cm squared for error. When writing out this differential should I replace 8 cm with x cm? So that it can be solved with any measurement? Is that what they are asking? Something to help me along would be much appreciated. Thanks.