Least Expensive!

Let the length and breadth of the tank be $x$ and $y$ meters respectively. It is given that the volume of the tank is $8m^3$ and height is $2m.$

$\large \displaystyle \therefore 2xy = 8 \implies xy= 4 \implies y = \frac{4}{x}\\ \text{Let C be the cost of the tank. Then }\\ \large \displaystyle C = 70xy + 45(2 \times 2y + 2 \times 2x) = 70xy + 180y + 180x\\ \large \displaystyle \implies C = 280 + \frac{720}{x} + 180x\\ \text{By differentiating with respect to x }\\ \large \displaystyle \implies \frac{dC}{dx} = -\frac{720}{x^2} + 180 \text{And } \frac{d^2C}{dx^2} = \frac{1440}{x^3}\\ \text{To find critical point, }\\ \large \displaystyle \therefore \frac{dC}{dx} = 0 \implies 180 - \frac{720}{x^2} = 0 \implies x = 2.\\ \large \displaystyle \text{Since, } \left(\frac{d^2C}{dx^2} \right)_{x=2} >0. \text{C is minimum at } x = 2.\\ \large \displaystyle \text{By putting } x= 2 \text{in the equation C, We get } C = 280 + \frac{720}{2} + 180 \times (2)\\ \large \displaystyle = 280 + 360 + 360 = \color{#D61F06}{\boxed{1000}}.$

$\large \displaystyle \therefore \text{The minimum cost for construction of tank is } \color{#20A900}{\boxed{\$1000}}.$