Step 1: Understanding the Concept
Let the width be \(x\) m and the length \(y\) m. Then \(xy=32\).
Step 2: Printed Area
Margins: top and bottom \(0.5\) m each, sides \(0.25\) m each.
Printed area \(P=(x-0.5)(y-1)=xy-x-0.5y+0.5\).
Step 3: Express in One Variable
With \(y=\dfrac{32}{x}\): \(P=32-x-\dfrac{16}{x}+0.5\).
\(\dfrac{dP}{dx}=-1+\dfrac{16}{x^{2}}=0\Rightarrow x=4\).
Step 4: Check and Conclude
\(\dfrac{d^{2}P}{dx^{2}}=-\dfrac{32}{x^{3}}<0\), so it is a maximum.
Then \(y=\dfrac{32}{4}=8\). The dimensions are 8 m and 4 m.
Final Answer:
The paper is 8 meters by 4 meters.
\[ \boxed{\text{(C) }8 \text{ meters and } 4 \text{ meters}} \]