Respuesta :
Let's draw a diagram of this problem.
ABC is the shore.
D to A is 1 miles (given).
A to C is 4 miles (given).
If we let AB = x, then BC would be "4 - x".
Now, using pythgorean theorem, let's find BD:
[tex]\begin{gathered} AB^2+AD^2=BD^2 \\ x^2+1^2=BD^2 \\ BD=\sqrt[]{1+x^2} \end{gathered}[/tex]We know
[tex]D=RT[/tex]Where
D is distance
R is rate
T is time
Swimmer needs to go from D to B at 3 miles per hour. Thus, we can say:
[tex]\begin{gathered} D=RT \\ T=\frac{D}{R} \\ T=\frac{\sqrt[]{1+x^2}}{3} \end{gathered}[/tex]Next part, swimmer needs to go from B to C at 6 miles per hour. Thus, we can say:
[tex]\begin{gathered} D=RT \\ T=\frac{D}{R} \\ T=\frac{4-x}{6} \end{gathered}[/tex]So, total time would be:
[tex]T=\frac{\sqrt[]{1+x^2}}{3}+\frac{4-x}{6}[/tex]We want to find the shortest possible time. From calculus we know that to find the shortest possible time, we need to differentiate the function T, set it equal to 0 to find the critical points and then use that point in the function T to find the shortest possible time.
Let's differentiate the function T:
[tex]\begin{gathered} T=\frac{\sqrt[]{1+x^2}}{3}+\frac{4-x}{6} \\ T=\frac{1}{3}(1+x^2)^{\frac{1}{2}}+\frac{4}{6}-\frac{1}{6}x \\ T=\frac{1}{3}(1+x^2)^{\frac{1}{2}}+\frac{2}{3}-\frac{1}{6}x \\ T^{\prime}=(\frac{1}{2})\frac{1}{3}(1+x^2)^{-\frac{1}{2}}\lbrack\frac{d}{dx}(1+x^2)\rbrack-\frac{1}{6} \\ T^{\prime}=\frac{1}{6}(1+x^2)^{-\frac{1}{2}}(2x)-\frac{1}{6} \\ T^{\prime}=\frac{2x}{6(1+x^2)^{\frac{1}{2}}}-\frac{1}{6} \\ T^{\prime}=\frac{x}{3\sqrt[]{1+x^2}}-\frac{1}{6} \end{gathered}[/tex]Now, we find the critical point:
[tex]\begin{gathered} T^{\prime}=\frac{x}{3\sqrt[]{1+x^2}}-\frac{1}{6} \\ T^{\prime}=0 \\ \frac{x}{3\sqrt[]{1+x^2}}-\frac{1}{6}=0 \\ \frac{x}{3\sqrt[]{1+x^2}}=\frac{1}{6} \\ \text{Cross Multiplying:} \\ 6x=3\sqrt[]{1+x^2} \\ \text{Square both sides:} \\ (6x)^2=(3\sqrt[]{1+x^2})^2 \\ 36x^2=9(1+x^2) \\ 36x^2=9+9x^2 \\ 36x^2-9x^2=9 \\ 27x^2=9 \\ x^2=\frac{9}{27} \\ x=\frac{\sqrt[]{9}}{\sqrt[]{27}} \\ x=\frac{3}{3\sqrt[]{3}} \\ x=\frac{1}{\sqrt[]{3}} \end{gathered}[/tex]Plugging this value into the equation of T, we get:
[tex]\begin{gathered} T=\frac{\sqrt[]{1+x^2}}{3}+\frac{4-x}{6} \\ T=\frac{\sqrt[]{1+(\frac{1}{\sqrt[]{3}})^2}}{3}+\frac{4-\frac{1}{\sqrt[]{3}}}{6} \\ T=\frac{\sqrt[]{1+\frac{1}{3}}}{3}+\frac{4-\frac{1}{\sqrt[]{3}}}{6} \\ T=\frac{\sqrt[]{\frac{4}{3}}}{3}+\frac{4-\frac{1}{\sqrt[]{3}}}{6} \\ T=\frac{\frac{2}{\sqrt[]{3}}}{3}+\frac{4-\frac{1}{\sqrt[]{3}}}{6} \\ T=\frac{2}{3\sqrt[]{3}}+\frac{4-\frac{1}{\sqrt[]{3}}}{6} \end{gathered}[/tex]Now, we can use the calculator to find the approximate value of T to be:
T = 0.9553 hours
This is the optimized time.
Converting to approximate minutes, it will be:
57.32 minutes
Answer:[tex]T=0.9553\text{ hours}[/tex]
