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close to 1, but the approximationusually gets worse as we move further away from 1.3. (3 pts) The edge of a cube was measured to be 8 cm, with a maximum possible error of 0.5 cm. Use a differentialto estimate the maximum possible error in computing the volume of the cube.(This next part is ungraded, just for fun. Using a calculator, find the actual error in measuring volume if theradius was really 8.5 cm instead of 8 cm, and find the actual error if the radius was actually 7.5 cm instead of 8cm. Compare these errors to the answer you got using differentials).Solution:V=x3, sodV= 3x2dx. Here,dxrepresents the error in the measurement (the difference betweenthe actual edge length and our measured edge length), anddVrepresents the corresponding approximate errorin volume.dV= 3·82·0.5 = 96 cm3.What was theactualerror in volume if we were off by 0.5 cm? It depends whether if 8 cm was an overestimateor underestimate.If the actual radius was 7.5 cm, thenV= 7.53= 421.88 cm3, while our approximation wasV= 83= 512cm3. The actual error would be 90.12 cm3.If the actual radius was 8.5 cm, thenV= 8.53= 614.13 cm3, while our approximation wasV= 512 cm3.The actual error would be 102.13 cm3.
4. (4 points) Alfador produces gear for a racquetball league. It costs him $0.005 per square centimeter to produceracquetballs for the league. In the past, the league used racquetballs with diameter 4 cm, and he produced 30,000racquetballs. However, league regulations now require racquetballs to have diameter 4.05 cm.