(Refer to picture for diagram.)
\(c=\sqrt{x^{2}+50^{2}}=\sqrt{x^{2}+2500}\)
\(Time=\frac{Distance}{Speed}\)
Time across river \(=\frac{\sqrt{x^{2}+2500}}{1.5}\)
Time along coast \(=\frac{200-x}{4}\)
So time \(T(x)=\frac{\sqrt{x^{2}+2500}}{1.5}+\frac{200-x}{4}\)
Differentiate: \(T'(x)=\frac{1}{1.5}*\frac{1}{2\sqrt{x^{2}+2500}}*2x+\frac{-1}{4}\)
⇒\(T'(x)=\frac{x}{1.5\sqrt{x^{2}+2500}}-\frac{1}{4}\)
For minimum, set \(T'(x)=0\)
⇒\(\frac{x}{1.5\sqrt{x^{2}+2500}}-\frac{1}{4}=0\)
⇒\(\frac{x}{1.5\sqrt{x^{2}+2500}}=\frac{1}{4}\)
⇒\(4x=1.5\sqrt{x^{2}+2500}\)
⇒\(16x^{2}=2.25(x^{2}+2500)\)
⇒\(16x^{2}=2.25x^{2}+5625\)
⇒\(13.75x^{2}=5625\)
⇒\(x^{2}=\frac{5625}{13.75}\)
⇒\(x=\sqrt{\frac{5625}{13.75}}\)
To avoid rounding errors, just plug the expression into the formula:
\(T=\frac{\sqrt{(\sqrt{\frac{5625}{13.75}})^{2}+2500}}{1.5}+\frac{200-\sqrt{\frac{5625}{13.75}}}{4}\)
T≈80.90 seconds
