1. **Problem statement:** We need to find the least positive integer $m$ such that the temperature $P(m)$ of the ingot is below 60°C. 2. **Given function:** $$P(x) = 22 + 68e^{-0.06x}, \quad x \geq 0$$ 3. **Condition to solve:** $$P(m) < 60$$ 4. **Substitute $P(m)$:** $$22 + 68e^{-0.06m} < 60$$ 5. **Isolate the exponential term:** $$68e^{-0.06m} < 60 - 22$$ $$68e^{-0.06m} < 38$$ 6. **Divide both sides by 68:** $$e^{-0.06m} < \frac{38}{68}$$ 7. **Simplify the fraction:** $$e^{-0.06m} < \frac{19}{34}$$ 8. **Take the natural logarithm of both sides:** $$\ln\left(e^{-0.06m}\right) < \ln\left(\frac{19}{34}\right)$$ 9. **Use the logarithm power rule:** $$-0.06m < \ln\left(\frac{19}{34}\right)$$ 10. **Divide both sides by -0.06, remembering to reverse the inequality because dividing by a negative number:** $$m > \frac{\ln\left(\frac{19}{34}\right)}{-0.06}$$ 11. **Calculate the right side:** $$\ln\left(\frac{19}{34}\right) = \ln(0.5588) \approx -0.581$$ $$m > \frac{-0.581}{-0.06} = 9.683$$ 12. **Since $m$ must be a positive integer, the least possible value is:** $$m = 10$$ **Final answer:** The least possible value of $m$ such that the ingot temperature is below 60°C is $\boxed{10}$ minutes.