I have been presented with the following question:
An open metal tank with a square base is made from $12 m^2$ of sheet metal. Find the length of the side of the base for the volume of the tank to be a maximum and find this maximum volume.
I have no idea how to answer this question, any help with the methodology showing how to get to the answer would be very much appreciated.
this is the original question:
Let the side of the square base be x m.
Then area of the square base is $x^2 m^2$
Area of metal sheet $12 m^2$
So out of $12 m^2$ we have $x^2$ used for base.
Then remaining area is used for 4 walls of container is $(12 −x^2) m^2$
This would be divided in the four sides/walls of the container with side x.
So we have $\frac{(12 −x^2) m^2}{4x m}$ as a height.
The volume of the container would be length * breadth * height.
V = $ x m * x m * \frac{(12 −x^2) m}{4x}$
= $\frac{(12x −x^3) m^3}{4}$
Now differentiate w.r.t x,
$\frac{d}{dx}V(x) = V′(x) = \frac{12 - 3x^2}{4} $
= $ 3 − \frac{3 x^2}{4}$
We equate it to 0 and get $x = \pm 2$
But x=2 becasue it can’t be −2 here.
Also,
$V′′(x) = \frac{−3x}{2} < 0$
So maximum volume.
Length of side x = 2 m, V(x) = $4 m^3$, height = 1 m.