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