1. **State the problem:** We have two vertical poles, one 16 ft high and the other 24 ft high, standing 30 ft apart. A worker wants to place a stake on the ground between them and run ropes from the stake to the top of each pole. We need to find the position of the stake that minimizes the total length of rope used. 2. **Set up variables:** Let the distance from the stake to the base of the 16 ft pole be $x$ feet. Then the distance from the stake to the base of the 24 ft pole is $30 - x$ feet. 3. **Express rope lengths:** The length of rope to the 16 ft pole is the hypotenuse of a right triangle with legs $x$ and 16, so $$L_1 = \sqrt{x^2 + 16^2} = \sqrt{x^2 + 256}.$$ Similarly, the length of rope to the 24 ft pole is $$L_2 = \sqrt{(30 - x)^2 + 24^2} = \sqrt{(30 - x)^2 + 576}.$$ 4. **Total rope length:** $$L = L_1 + L_2 = \sqrt{x^2 + 256} + \sqrt{(30 - x)^2 + 576}.$$ 5. **Minimize total length:** To find the minimum, differentiate $L$ with respect to $x$ and set derivative to zero: $$\frac{dL}{dx} = \frac{x}{\sqrt{x^2 + 256}} - \frac{30 - x}{\sqrt{(30 - x)^2 + 576}} = 0.$$ 6. **Solve for $x$:** $$\frac{x}{\sqrt{x^2 + 256}} = \frac{30 - x}{\sqrt{(30 - x)^2 + 576}}.$$ Cross-multiplied: $$x \sqrt{(30 - x)^2 + 576} = (30 - x) \sqrt{x^2 + 256}.$$ 7. **Square both sides:** $$x^2 ((30 - x)^2 + 576) = (30 - x)^2 (x^2 + 256).$$ Expand: $$x^2 (900 - 60x + x^2 + 576) = (900 - 60x + x^2)(x^2 + 256).$$ Simplify inside parentheses: $$x^2 (x^2 - 60x + 1476) = (x^2 - 60x + 900)(x^2 + 256).$$ 8. **Expand both sides:** Left: $$x^4 - 60x^3 + 1476x^2.$$ Right: $$x^4 - 60x^3 + 900x^2 + 256x^2 - 15360x + 230400 = x^4 - 60x^3 + 1156x^2 - 15360x + 230400.$$ 9. **Bring all terms to one side:** $$x^4 - 60x^3 + 1476x^2 - (x^4 - 60x^3 + 1156x^2 - 15360x + 230400) = 0,$$ which simplifies to $$1476x^2 - 1156x^2 + 15360x - 230400 = 0,$$ $$320x^2 + 15360x - 230400 = 0.$$ 10. **Divide entire equation by 320:** $$x^2 + 48x - 720 = 0.$$ 11. **Solve quadratic:** $$x = \frac{-48 \pm \sqrt{48^2 - 4 \times 1 \times (-720)}}{2} = \frac{-48 \pm \sqrt{2304 + 2880}}{2} = \frac{-48 \pm \sqrt{5184}}{2} = \frac{-48 \pm 72}{2}.$$ 12. **Two solutions:** $$x = \frac{-48 + 72}{2} = \frac{24}{2} = 12,$$ $$x = \frac{-48 - 72}{2} = \frac{-120}{2} = -60.$$ Since $x$ is a distance along the ground between the poles, it must be between 0 and 30, so $x = 12$ ft. 13. **Answer:** The stake should be placed 12 feet from the base of the 16 ft pole to minimize the total rope length.