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The part of the diagram inside the wrapped film is a $w$ by $d$ rectangle and two semicircles of radius $r$ and hence its area is \[w \times d + \frac{\pi}{4} w^2 \mbox{ square inches.}\] If $W$ is the width of the wrapped spool as in the diagram then the area of the spool including wrapped film and the region inside it is \[W \times d + \frac{\pi}{4} W^2 \mbox{ square inches.}\] The area of the edge of the wrapped spool of film, the race track shaped blue region in the diagram is thus \[W \times d + \frac{\pi}{4} W^2  w \times d + \frac{\pi}{4} w^2 \mbox{ square inches.}\] This simplifies to \[ d(Ww) + \frac{\pi}{4} \left(w^2  w^2 \right) \mbox{ square inches.}\] If the film is of length $L$ inches and its thickness is $t$ inches then one edge of the film can be seen as a long thin rectangle, $L$ by $t$ inches. The area of this rectangle is $t \times L$ square inches. When rolled this long edge forms the blue, race track shaped region in the diagram and hence \[ tL = d(Ww) + \frac{\pi}{4} \left(W^2  w^2 \right) \mbox{ square inches.}\] Rewriting this equation I got \[\frac{\pi}{4} W^2 + dW  dw \frac{\pi}{4} w^2  tL = 0.\] This is a quadratic equation in $W$ and the general quadratic equation gives \[W = \frac{2 \left( d \pm \sqrt{d^2  \pi \; \left( dw + \frac{w^2}{4} \pi \; + tL \right) } \right)}{\pi} \mbox{ inches.}\] Choosing the plus sign in front of the square root symbol gives the width with the film wrapped on, and from the diagram the length is \[d + W \mbox{ inches.}\] Harley




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