The edge of a cube was measured to be 8 cm, with a maximum possible error of 0.5 cm. Use a differential to 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 the radius was really 8.5 cm instead of 8 cm, and find the actual error if the radius was actually 7.5 cm instead of 8 cm. Compare these errors to the answer you got using differentials).

Answer: 96 cc102.125 cc90.125 ccStep-by-step explanation:Let V(x) the volume of a cube of edge x cm. Then[tex] \bf V(x)=x^3 \;cm^3[/tex]The maximum possible error in computing the volume of the cube would beV(8+0.5) - V(8)By approximating with the derivative, we knowV(8+0.5) ≅ V(8) +0.5V'(8)henceV(8+0.5) - V(8) ≅ 0.5V'(8)  But[tex] \bf V'(x)=3x^2\Rightarrow V'(8)=3*64=192 [/tex]and maximum possible error in computing the volume of the cube would be approximately 0.5*192 = 96 cc.find the actual error in measuring volume if the radius was really 8.5 cm instead of 8 cm[tex] \bf V(8.5) - V(8) = (8.5)^3-8^3=614.125-512=102.125\;cm^3[/tex]find the actual error if the radius was actually 7.5 cm instead of 8 cm.[tex] \bf V(8) - V(7.5) = (8)^3-(7.5)^3=512-421.875=90.125\;cm^3[/tex]