Hello, LLemon!

This is __not__ a simple problem . . .

A fish tank has a triangular cross-section.

It has a length of 80cm, width of 50cm and a depth of 40cm.

Initially it is full of water.

After 4 days, the depth of the water had dropped to 35cm.

The rate of evaporation is proportional to the surface area of the water.

Calculate how long the water in the fish tank takes to completely evaporatel. Code:

25
- *-------*-------*
: \ | /
: \ | x /
: \- - + - -/
40 \:::|:::/
: \::|h:/
: \:|:/
: \|/
- *

The volume of water is: .$\displaystyle V \;=\;\frac{1}{2}(2x)(h)(80) \;=\;80hx$ .**[1]**

From the similar right triangles: .$\displaystyle \frac{x}{h} \,=\,\frac{25}{40}\quad\Rightarrow\quad x \:=\:\frac{5}{8}\,h$ .**[2]**

Substitute [2] into [1]: .$\displaystyle V \;=\;80h\left(\frac{5}{8}h\right) \;=\;50h^2$

. . Then: .$\displaystyle \frac{dV}{dt} \:=\:100h\left(\frac{dh}{dt}\right)$ .**[3]**

We are told that $\displaystyle \frac{dV}{dt}$ is proportional to the surface area: .$\displaystyle \frac{dV}{dt} \:=\:k\!\cdot\!A$

. . The surface area is: .$\displaystyle A \:=\:(2x)(80) \:=\: 160x$

. . Subtitute [2]: .$\displaystyle A \:=\:160\left(\frac{5}{8}h\right) \:=\:100h$

Then: .$\displaystyle \frac{dV}{dt} \;=\;k(100h) \;=\;

100kh$ .**[4]**

Equate [3] and [4]: .$\displaystyle 100h\left(\frac{dh}{dt}\right)\;=\;100kh \quad\Rightarrow\quad dh \:=\:k\,dt$

Integrate: .$\displaystyle h \;=\;kt + C$

When $\displaystyle t = 0,\:h = 40\!:\;\;40 \:=\:k(0) + C\quad\Rightarrow\quad C = 40$

. . The height function (so far) is: .$\displaystyle h \;=\;kt + 40$

When $\displaystyle t = 4,\:h = 35\!:\;\;35\:=\:k(4) + 40\quad\Rightarrow\quad k \,=\,-\frac{5}{4}$

Hence, the height function is: .$\displaystyle h \;=\;-\frac{5}{4}\,t + 40$

When is the tank empty $\displaystyle (h = 0)$?

. . $\displaystyle -\frac{5}{4}\,t + 40 \:=\:0\quad\Rightarrow\quad t \:=\:32$

Therefore, it takes $\displaystyle \boxed{{\color{blue}\,32\text{ hours}\,}}$.