Peter constructed an open fish tank with a rectangular base of length $4l$ m, breadth $l$ m, and height $h$ m. He wanted the total outer surface area of the fish tank to be $5\text{ m}^2$.

1. Formulate the equation for the total outer surface area of the open tank.

Area of base = $4l \times l = 4l^2$
Area of 4 vertical sides = $2(4l \times h) + 2(l \times h) = 8lh + 2lh = 10lh$
Total surface area = $4l^2 + 10lh$
Given total surface area is $5$, so:
$4l^2 + 10lh = 5$

2. Express the height $h$ in terms of $l$.

$10lh = 5 - 4l^2$
$h = \frac{5 - 4l^2}{10l}$

3. Substitute $h$ into the volume formula and simplify to show the given expression.

Volume $V = \text{base area} \times \text{height}$
$V = 4l^2 \times h$
$V = 4l^2 \left( \frac{5 - 4l^2}{10l} \right)$
$V = \frac{4l^2}{10l} (5 - 4l^2)$
$V = \frac{2l}{5} (5 - 4l^2)$
$V = \frac{2}{5}(5l - 4l^3)$ (shown)

1. Differentiate the volume function with respect to $l$

$V = \frac{2}{5}(5l - 4l^3) = 2l - \frac{8}{5}l^3$

$\frac{dV}{dl} = 2 - \frac{24}{5}l^2$

2. Set the derivative to zero to find the stationary point

For maximum volume, $\frac{dV}{dl} = 0$

$2 - \frac{24}{5}l^2 = 0$

$\frac{24}{5}l^2 = 2$

$l^2 = 2 \times \frac{5}{24} = \frac{5}{12}$

3. Verify that this value of $l$ gives a maximum volume

$\frac{d^2V}{dl^2} = -\frac{48}{5}l$

Since $l > 0$ (length must be positive), $\frac{d^2V}{dl^2} < 0$.

Therefore, the volume is maximum when $l^2 = \frac{5}{12}$.

4. Calculate the area of the rectangular base

Area of base = $4l \times l = 4l^2$

Area = $4 \left(\frac{5}{12}\right) = \frac{5}{3}\text{ m}^2$