Question 1161726
b) You can stamp up to {{{highlight(40)}}} circles,
Stamping the 60cm diameter circles centered on the mid-width of the sheet, 25mm (2.5cm) apart from each other would allow us to stamp along the 25m (2500cm) steel coil
{{{2500cm/(60cm+2.5cm)=40}}}circles.
In fact {{{40}}} circles with {{{40-1=39}}} gaps between them, adding {{{2.5cm}}} per gaps adds to a length of
{{{40*(60cm)+39*(2.5cm)=2497.5cm}}}
and we could use the {{{2500cm-2497.5cm=2.5cm}}} leftover length
to leave some space between the circles at the ends and the beginning and end of the coil.
Offsetting the circles so they are shifted a bit to the left and the right, makes the string of circles a little shorter, but not short enough to fit an additional circle.
The most you could offset centers is {{{70cm-60cm=10cm}}} ,
and making the distance between the centers centers {{{AB=62.5cm}}} , would make the circles look like this
{{{drawing(650,400,-5,125,-40,40,
red(rectangle(0,-35,121.7,35)),rectangle(121.7,-35,150,35),
red(circle(30,-5,30)),red(circle(30,-5,0.4)),red(circle(30,-5,0.2)),
red(circle(91.7,5,30)),red(circle(91.7,5,0.4)),red(circle(91.7,5,0.2)),
triangle(30,-5,91.7,-5,91.7,5),rectangle(91.8,-5,90.7,-4),
locate(28.5,-5.5,A),locate(91,8.5,B),locate(91,-5.1,C),
red(circle(151.7,-5,30)),arrow(0,-5,30,-5),arrow(30,-5,0,-5),
arrow(91.7,5,121.7,5),arrow(121.7,5,91.7,5),arrow(100,0,100,-35),
arrow(100,0,100,35),locate(13,-5,30cm),locate(104.7,5,30cm),
locate(101,0,70cm)
)}}} with {{{BC=10cm}}} and {{{AC=sqrt(62.5^2+10^2)}}}{{{cm=61.7cm}}} .
Each pattern with 2 circles (red rectangle) takes up
{{{30cm+61.7cm+30cm=121.7cm}}} , so {{{40}}} circles require {{{40+121.7=2434cm}}} out of the {{{2500cm}}} length of the coil.
 
a) The area of each circle is {{{pi*radius^2=pi*(30cm)^2=900pi}}}{{{cm^2}}}
The area of the square cut off is {{{(30cm)*(30cm)=900cm^2}}} .
The area of each component stamped is
{{{(900pi-900)}}}{{{cm^2=900(pi-1)}}}{{{cm^2=approx.}}}{{{900(3.1416-1)=900*2.1416cm^2=1927.44cm^2}}}
The total area of the {{{40}}} stamped components is
{{{40*1927.44cm^2=77097.6cm^2}}}
The area of a sheet coil {{{70cm}}} wide and {{{2500cm}}} long is
{{{(70cm)*(2500cm)=175000cm^2}}} .
Out of that surface area {{{77097.6cm^2}}} turns into {{{40}}} components, and the rest is "steel wasted including the center square".
The surface area of the steel wasted including the center square" is
{{{175000cm^2-77097.6cm^2=97902.4cm^2}}}
As a percentage of the {{{175000cm^2}}} in the coil, that is
{{{(97902.4/175000cm^2)*"100%"="approx."}}}{{{highlight("55.9%")="approx."}}}{{{highlight("56%")}}}
 
c) The {{{"56%"=56/100=0.56}}} of the coil wasted represents a cost of {{{0.56*"$200"="$112"}}} .
Selling it at {{{"50%"=50/100=0.5}}} of the cost reclaims {{{0.5*"$112"="$56"}}} ,
making the cost of {{{40}}} components {{{"$200"-"$56"-"$144"}}} , and the cost per component
{{{"$144"/40=highlight("$3.60")}}}