10.6. The sides of the base of a brick are 28 cm and 9 cm, and the height is 6 cm. A snail crawls in a straight line along the faces of the brick from one vertex of the lower base to the opposite vertex of the upper base. The horizontal and vertical components of its velocity, $v_{x}$ and $v_{y}$, are related by the equation $v_{x}^{2}+4 v_{y}^{2}=1$ (for example, on the top face $v_{y}=0$ cm/min, so $v_{x}=v=1$ cm/min). What is the minimum time the snail can spend on its journey?
Answer: 35 min.

Fig. $10.6 \mathrm{a}$

Solution. Let the given brick be a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D$, where $A B-=9 \mathrm{~cm}, B C=28 \mathrm{~cm}, A A^{\prime}=6 \mathrm{~cm}$.

First method. Increase the height of the brick and the vertical component of the snail's speed by two times. We get a brick $A B C D E F G H$, on which the snail's speed will always be the same: $v=\sqrt{V_{x}^{2}+\left(2 V_{y}\right)^{2}}=1 \mathrm{~cm} /$ min, and the minimum time of movement is the same as on the original brick.

Suppose the snail crawls from vertex $B$ to vertex $F$. Consider the development of three faces of this brick: the base $A B C D$ and two adjacent vertical faces with a common edge $D F$. Since the snail's speed is constant, the minimum time will be when the path length is the shortest, that is, when moving along the diagonal of one of the two rectangles: BCFE or $A B G F$ (see Fig. 10.6a). By the Pythagorean theorem, we get that $B F=35$ cm; $B F=41$ cm. Therefore, it is more advantageous to move along the diagonal $B F$, then at a speed of 1 cm/min, the minimum time will be 35 minutes.

Second method. Notice that the snail must crawl 6 cm vertically in any case, so we can assume that it uniformly crawled from vertex $D$ to vertex $B'$ and has already reached the top face of the brick. In this case, it ended up at one of the points $A_{1}$ or $A_{2}$ at a distance $a$ from vertex $D_{1}$ (see Fig. 10.6b).

![](https://cdn.mathpix.com/cropped/2024_05_06_f19840638bf8a4639daeg-4.jpg?height=323&width=788&top_left_y=495&top_left_x=1134)

Let it have spent $t$ minutes on this path, then multiplying both sides of the equation $v_{x}^{2}+4 v_{y}^{2}=1$ by $t^{2}$, we get: $t^{2} v_{x}^{2}+4 t^{2} v_{y}^{2}=t^{2}$. In this case, $a=t v_{x}$ and $6=t v_{y}$, so $a^{2}+4 \cdot 36=t^{2}$, from which $t=\sqrt{a^{2}+144}$.

Consider two possible routes on the top face: $A_{1} B^{\prime}=\sqrt{28^{2}+\left(9-a_{1}\right)^{2}}$ or $A_{2} B^{\prime}=\sqrt{9^{2}+\left(28-a_{2}\right)^{2}}$. The speed of its movement on the top face is 1 cm/min, so the time is numerically equal to the length of the path traveled. Find the values of $a$ for which the functions $f_{1}(a)=\sqrt{a^{2}+144}+\sqrt{28^{2}+(9-a)^{2}}$ and $f_{2}(a)=\sqrt{a^{2}+144}+\sqrt{9^{2}+(28-a)^{2}}$ take their minimum values.

For the first function, the geometric interpretation is the sum of the distances on the coordinate plane between the points $(0 ; 0)$ and $(12 ; a)$ and between the points $(12 ; a)$ and $(40 ; 9)$. This sum takes the minimum value if the point $(12 ; a)$ lies on the segment with endpoints at the other two points. Therefore, $\frac{40}{12}=\frac{9}{a}$, that is, $a_{1}=2.7$, $f_{1}(2.7)=\sqrt{2.7^{2}+144}+\sqrt{28^{2}+6.3^{2}}=12.3+28.7=41$.

For the second function, the geometric interpretation is the sum of the distances between the points $(0 ; 0)$ and $(12 ; a)$ and between the points $(12 ; a)$ and $(21 ; 28)$. Reasoning similarly, we get: $\frac{21}{12}=\frac{28}{a}$, then $a_{2}=16, f_{2}(16)=\sqrt{16^{2}+144}+\sqrt{9^{2}+12^{2}}=20+15=35$.

Therefore, the minimum time of movement is 35 minutes.

Students who already know how to compute derivatives of functions can act differently in the final part of the solution.

Find the derivatives of the obtained functions:

$$
f_{1}^{\prime}(a)=\frac{a}{\sqrt{a^{2}+144}}+\frac{a-9}{\sqrt{28^{2}+(9-a)^{2}}} u
$$

$f_{2}^{\prime}(a)=\frac{a}{\sqrt{a^{2}+144}}+\frac{a-28}{\sqrt{9^{2}+(28-a)^{2}}} . \quad$ Zeros of the derivatives: $a_{1}=2.7 ; \quad a_{2}=16 . \quad$ Since the derivatives take negative values to the left of these values of $a$ and positive values to the right, these points are the only points of minimum of the functions. Due to the continuity of each function on its domain, the minimum values of these functions will be achieved at these points. These values are: $f_{1}(2.7)=\sqrt{2.7^{2}+144}+\sqrt{28^{2}+6.3^{2}}=12.3+28.7=41$ and $f_{2}(16)=\sqrt{16^{2}+144}+\sqrt{9^{2}+12^{2}}=20+15=35$.

Grading criteria. «+» A complete and justified solution is provided

«士» The correct answer is obtained justifiably, based on the consideration of only one movement option

«Ғ» The answer 41 minutes is obtained justifiably

«Ғ» The time functions are obtained justifiably, but there is no further progress or it contains errors

«-» Only the answer is provided

«-» The problem is not solved or is solved incorrectly