A laboratory wants to build two tanks, a closed metal tank, another tank without a side face, both with a square bottom, the first must have a capacity of 40m^3 and the second 55m^3. What dimensions should both tanks have so that the least amount of metal is needed in their manufacture?

1. Determine the dimensions of the closed metal deposit (with a square base) to minimize the surface area:

Given the volume of \( V_1 = 40 \, \text{m}^3 \):

Let \( a_1 \) be the side length of the square base, and \( h_1 \) be the height of the deposit.

V_1 = a_1^2 \cdot h_1 \Rightarrow a_1^2 \cdot h_1 = 40 \Rightarrow h_1 = \frac{40}{a_1^2}

Surface area \( S_1 \) to be minimized (including all six faces: base, top, and four sides):

S_1 = a_1^2 + 4a_1 h_1 + a_1^2 = 2a_1^2 + 4a_1 \cdot \frac{40}{a_1^2} = 2a_1^2 + \frac{160}{a_1}

To minimize \( S_1 \), take the derivative and set it to zero:

\frac{dS_1}{da_1} = 4a_1 - \frac{160}{a_1^2} = 0

Solving for \(a_1\):

4a_1^3 = 160 \Rightarrow a_1^3 = 40 \Rightarrow a_1 = \sqrt[3]{40}

Thus, the height:

h_1 = \frac{40}{(\sqrt[3]{40})^2} = \sqrt[3]{40}

Therefore, dimensions of the first deposit:

a_1 = \sqrt[3]{40}, \quad h_1 = \sqrt[3]{40}

2. Determine the dimensions of the second deposit (with one open face, square base) to minimize surface area:

Given volume \( V_2 = 55 \, \text{m}^3 \):

Let \( a_2 \) be the side length of the square base, and \( h_2 \) be the height of the deposit.

V_2 = a_2^2 \cdot h_2 \Rightarrow a_2^2 \cdot h_2 = 55 \Rightarrow h_2 = \frac{55}{a_2^2}

Surface area \( S_2 \) to minimize (including the base and four sides):

S_2=2a_2^2+3a_2h_2=2a_2^2+3a_2\cdot\frac{55}{a_2^2}=2a_2^2+\frac{165}{a_2}

To minimize \( S_2 \), take the derivative and set it to zero:

\frac{dS_2}{da_2}=4a_2-\frac{165}{a_2^2}=0

Solving for \( a_2 \):

4a_2^3=165\Rightarrow a_2^3=\frac{165}{4}\Rightarrow a_2=\sqrt[3]{41.25}

Thus, the height:

h_2=\frac{55}{(\sqrt[3]{41.25})^2}=\frac{55}{\sqrt[3]{1701.5625}}=\sqrt[3]{97.778}

Therefore, dimensions of the second deposit:

a_2=\sqrt[3]{41.25},\quad h_2=\left(97.78\right)^{\left(\frac{1}{3}\right)}