a fence 10 ft high runs parallel to a tall building at a distance of 6 ft from the building. what is the…

Explanation:

Step1: Set up the geometric - ratio relationship

Let the distance from the base of the ladder to the fence be $x$ feet. Let the length of the ladder be $L$. Using similar right - triangles, the height $y$ on the building that the ladder reaches is given by $y=\frac{10(x + 6)}{x}$. By the Pythagorean theorem, $L^{2}=(x + 6)^{2}+y^{2}=(x + 6)^{2}+\left(\frac{10(x + 6)}{x}\right)^{2}=(x + 6)^{2}\left(1+\frac{100}{x^{2}}\right)$.

Step2: Simplify the expression for $L^{2}$

Expand $(x + 6)^{2}=x^{2}+12x + 36$. Then $L^{2}=(x^{2}+12x + 36)\left(1+\frac{100}{x^{2}}\right)=x^{2}+12x + 36+\frac{100x^{2}}{x^{2}}+\frac{1200x}{x^{2}}+\frac{3600}{x^{2}}=x^{2}+12x + 136+\frac{1200}{x}+\frac{3600}{x^{2}}$.

Step3: Differentiate $L^{2}$ with respect to $x$

Let $u = L^{2}$. Then $u^\prime=2x+12-\frac{1200}{x^{2}}-\frac{7200}{x^{3}}$. Set $u^\prime = 0$ to find the critical points. Multiply through by $x^{3}$ to get $2x^{4}+12x^{3}-1200x - 7200 = 0$. Divide by 2: $x^{4}+6x^{3}-600x - 3600 = 0$. We can also use a more intuitive approach. Let the angle between the ladder and the ground be $\theta$. From similar triangles, $\tan\theta=\frac{10}{x}$ and the length of the ladder $L=\frac{10}{\sin\theta}+\frac{6}{\cos\theta}$. Differentiate $L$ with respect to $\theta$: $L^\prime=-\frac{10\cos\theta}{\sin^{2}\theta}+\frac{6\sin\theta}{\cos^{2}\theta}$. Set $L^\prime = 0$, then $\frac{10\cos\theta}{\sin^{2}\theta}=\frac{6\sin\theta}{\cos^{2}\theta}$, which simplifies to $10\cos^{3}\theta=6\sin^{3}\theta$, or $\tan^{3}\theta=\frac{10}{6}=\frac{5}{3}$, so $\tan\theta=\sqrt[3]{\frac{5}{3}}$. Since $\tan\theta=\frac{10}{x}=\sqrt[3]{\frac{5}{3}}$, then $x = 10\sqrt[3]{\frac{3}{5}}$.

Step4: Calculate the length of the ladder

$L=\frac{10}{\sin\theta}+\frac{6}{\cos\theta}$. Since $\tan\theta=\sqrt[3]{\frac{5}{3}}$, we know that $\sin\theta=\frac{\sqrt[3]{\frac{5}{3}}}{\sqrt{1 + (\sqrt[3]{\frac{5}{3}})^2}}$ and $\cos\theta=\frac{1}{\sqrt{1+(\sqrt[3]{\frac{5}{3}})^2}}$. Another way: From the similar - triangles approach, when we set $u^\prime = 0$ and solve $x^{4}+6x^{3}-600x - 3600 = 0$, we can also use a numerical method or observe that by trial and error or using a graphing utility, $x\approx4.48$. $L=\sqrt{(x + 6)^{2}+\left(\frac{10(x + 6)}{x}\right)^{2}}$. Substituting $x\approx4.48$ into the formula: $L=\sqrt{(4.48+6)^{2}+\left(\frac{10(4.48 + 6)}{4.48}\right)^{2}}=\sqrt{(10.48)^{2}+\left(\frac{10\times10.48}{4.48}\right)^{2}}=\sqrt{109.8304+\left(\frac{104.8}{4.48}\right)^{2}}=\sqrt{109.8304 + 547.396}=\sqrt{657.2264}\approx25.6$

Answer:

$25.6$