4.7 Applied optimization problems  (Page 3/8)

Step 1: Let $x$ be the distance running and let $y$ be the distance swimming ( [link] ). Let $T$ be the time it takes to get from the cabin to the island.

Step 2: The problem is to minimize $T.$

Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance $=$ Rate $\times$ Time $(D=R\times T),$ the time spent running is

and the time spent swimming is

Therefore, the total time spent traveling is

Step 4: From [link] , the line segment of $y$ miles forms the hypotenuse of a right triangle with legs of length $2\text{mi}$ and $6-x\text{mi}.$ Therefore, by the Pythagorean theorem, $2^2+{\left(6-x\right)}^2=y^2,$ and we obtain $y=\sqrt{{\left(6-x\right)}^2+4}.$ Thus, the total time spent traveling is given by the function

Step 5: From [link] , we see that $0\le x\le 6.$ Therefore, $\left[0,6\right]$ is the domain of consideration.

Step 6: Since $T\left(x\right)$ is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let’s begin by looking for any critical points of $T$ over the interval $\left[0,6\right].$ The derivative is

If $T^{\prime }\left(x\right)=0,$ then

Therefore,

Squaring both sides of this equation, we see that if $x$ satisfies this equation, then $x$ must satisfy

which implies

We conclude that if $x$ is a critical point, then $x$ satisfies

Therefore, the possibilities for critical points are

Since $x=6+6\text{/}\sqrt{55}$ is not in the domain, it is not a possibility for a critical point. On the other hand, $x=6-6\text{/}\sqrt{55}$ is in the domain. Since we squared both sides of [link] to arrive at the possible critical points, it remains to verify that $x=6-6\text{/}\sqrt{55}$ satisfies [link] . Since $x=6-6\text{/}\sqrt{55}$ does satisfy that equation, we conclude that $x=6-6\text{/}\sqrt{55}$ is a critical point, and it is the only one. To justify that the time is minimized for this value of $x,$ we just need to check the values of $T\left(x\right)$ at the endpoints $x=0$ and $x=6,$ and compare them with the value of $T\left(x\right)$ at the critical point $x=6-6\text{/}\sqrt{55}.$ We find that $T\left(0\right)\approx 2.108\text{h}$ and $T\left(6\right)\approx 1.417\text{h,}$ whereas $T\left(6-6\text{/}\sqrt{55}\right)\approx 1.368\text{h}.$ Therefore, we conclude that $T$ has a local minimum at $x\approx 5.19$ mi.