1. **Problem Statement:** We want to minimize the time function $T$ by finding the critical points where its derivative $\frac{dT}{dx}$ equals zero. 2. **Given Derivative Expressions:** $$\frac{dT}{dx} = \frac{2x}{12\sqrt{20^2 + x^2}} + \frac{-80 + 2x}{4\sqrt{2000 - 80x + x^2}}$$ 3. **Step 1: Simplify each term separately.** - First term: $$\frac{2x}{12\sqrt{400 + x^2}} = \frac{x}{6\sqrt{400 + x^2}}$$ - Second term: $$\frac{-80 + 2x}{4\sqrt{2000 - 80x + x^2}}$$ remains as is for now. 4. **Step 2: Set the derivative equal to zero to find critical points:** $$\frac{x}{6\sqrt{400 + x^2}} + \frac{-80 + 2x}{4\sqrt{2000 - 80x + x^2}} = 0$$ 5. **Step 3: Multiply both sides by the common denominator to clear fractions:** Multiply both sides by $$12 \sqrt{400 + x^2} \sqrt{2000 - 80x + x^2}$$ to get: $$2x \sqrt{2000 - 80x + x^2} + 3(-80 + 2x) \sqrt{400 + x^2} = 0$$ 6. **Step 4: Rearrange the equation:** $$2x \sqrt{2000 - 80x + x^2} = -3(-80 + 2x) \sqrt{400 + x^2}$$ 7. **Step 5: Square both sides to eliminate square roots:** $$4x^2 (2000 - 80x + x^2) = 9(-80 + 2x)^2 (400 + x^2)$$ 8. **Step 6: Expand and simplify both sides:** - Left side: $$4x^2 (2000 - 80x + x^2) = 8000x^2 - 320x^3 + 4x^4$$ - Right side: $$9(6400 - 320x + 4x^2)(400 + x^2)$$ 9. **Step 7: Expand the right side:** $$9[(6400)(400) + (6400)(x^2) - (320x)(400) - (320x)(x^2) + (4x^2)(400) + (4x^2)(x^2)]$$ $$= 9[2,560,000 + 6400x^2 - 128,000x - 320x^3 + 1600x^2 + 4x^4]$$ $$= 9[2,560,000 + 8000x^2 - 128,000x - 320x^3 + 4x^4]$$ 10. **Step 8: Multiply by 9:** $$23,040,000 + 72,000x^2 - 1,152,000x - 2,880x^3 + 36x^4$$ 11. **Step 9: Set the equation:** $$8000x^2 - 320x^3 + 4x^4 = 23,040,000 + 72,000x^2 - 1,152,000x - 2,880x^3 + 36x^4$$ 12. **Step 10: Bring all terms to one side:** $$0 = 23,040,000 + 72,000x^2 - 1,152,000x - 2,880x^3 + 36x^4 - 8000x^2 + 320x^3 - 4x^4$$ 13. **Step 11: Combine like terms:** $$0 = 23,040,000 - 1,152,000x + (72,000x^2 - 8,000x^2) + (-2,880x^3 + 320x^3) + (36x^4 - 4x^4)$$ $$0 = 23,040,000 - 1,152,000x + 64,000x^2 - 2,560x^3 + 32x^4$$ 14. **Step 12: This quartic equation can be solved numerically or graphically to find the value(s) of $x$ that minimize $T$.** 15. **Summary:** We derived the critical point condition by setting $\frac{dT}{dx} = 0$ and simplifying to a quartic polynomial: $$32x^4 - 2560x^3 + 64000x^2 - 1152000x + 23040000 = 0$$ Solving this equation will give the $x$ values that minimize the time $T$. **Final answer:** The critical points satisfy $$32x^4 - 2560x^3 + 64000x^2 - 1152000x + 23040000 = 0$$ You can use numerical methods or graphing tools to find the exact minimizing $x$.