Calculus Optimization problem part 2

Question

A person is sailing across a circular lake with diameter $4$ miles. He starts at point $X$ and needs to get to point $Y$, which is diametrically opposite to $X$. To get there as quickly as possible, he will sail directly from $X$ to a point $Z$ on the shore and then walk from $Z$ to $Y$ along the rim of the lake. If he can sail at a speed of $2$ miles per hour and can walk $4$ miles per hour, what is the minimum number of hours he needs to make his trip?

Any help would be great!

Answer 1

so you want to minimize the time so how about obtaining an expression of time with only one variable say $\theta$ see the diagram.

now time taken is $$\frac{XZ}{2}+\frac{r\theta}{4} \space\space hours$$

where $r\theta$ is the length of the curve $ZQY$

r is given 2 miles thus our expression becomes $$\frac{XZ}{2}+\frac{\theta}{2} \space\space hours$$

now we need to calculate the length of the path XZ apply cosine rule where The angle ZPX is $\pi -\theta$

$$cos(\pi - \theta)=\frac{r^2+r^2-(XZ)^2}{2rr}$$

$$XZ^2=8(1+Cos\theta)$$ $$XZ=2\sqrt{2(1+Cos\theta)}$$ thus the time taken is$$\sqrt{2(1+Cos\theta)}+\frac{\theta}{2} \space\space hours$$

differentiate it w.r.t. $\theta$ and set equal 0 gives $$cos\theta=-1,\frac{1}{2}$$ where $-1$ gives us minimum value of the time

thus the minimum time taken is $$\frac{\pi}{2} \space hours$$

"So, don't sail at all... walk the whole way along the semicircle.

Answer 2

Let's define: $X=[x_{x},x_{y}]=[-2,0]$, $Y=[y_{x},y_{y}]=[2,0]$ and $Z=[z_{x},z_{y}]$ be the initial, final and sail-walk change points in the plane, respectively.

If $Z$ belong to the circunference with radio 2 and center $[0,0]$. Then, $Z=[z_{x},(4-z_{x}^2)^{\frac{1}{2}}]$.

Thus, let's define a scalar function of time $T(z_{x})$ w.r.t. z_{x} coordinate.

If time = distance/velocity, then

$T(z_{x})=\frac{d_s}{v_s}+\frac{d_w}{v_w}$

is the sum of both, sailing and walking, velocities.

Sailing velocity $v_s=2$ and walking velocity $v_w=4$.

Then, the sailing distance can be computed as

$d_s=((x_x-z_x)^2+(x_y-(4-z_x^2)^{\frac{1}{2}})^2)^{\frac{1}{2}}$

and the walking distance as

$d_w=r\theta=2\tan^{-1}(\frac{z_y}{z_x})$

Remembering that $z_y=(4-z_{x}^2)^{\frac{1}{2}}$, we can define the functional of time w.r.t. $z_x$

$T(z_x)=\frac{1}{2}(8+4z_x)^{\frac{1}{2}}+\frac{1}{2}\tan^{-1}(\frac{(4-z_x^2)^{\frac{1}{2}}}{z_x})$

Now, the univariate unconstrained optimization problem is solved through

$T'(z_x)=0$

Differentiating

$T'(z_x)=\frac{1}{2}(2+z_x)^{-\frac{1}{2}}-\frac{1}{2}(4-z_x^2)^{-\frac{1}{2}}=0$

$T'(z_x)=z_x^2+z_x-2=0$

Then, solving for $z_x$ we found two stationary points

$z_{x1} = 1$ and $z_{x2}=-2$

The point $z_{x1} = 1$ is a maximum where the slope is zero.

The point $z_{x2}=-2$ is the absolute minimum in the [-2,2]-x-axis range.

Then the minumum time is 1.57 hours and is obtained when the $X$ and $Z$ points are equal, which means sealing is not feasible, so just walk instead.