Question

A rectangular metal tank with an open top is to hold 171.5 cubic feet of liquid. what are the dimensions of the tank that require the least material to build?

Answer 1

To minimize the material usage we have to have the volume requested with the minimum surface area.
The volume is:
$171.5 =xyz$
And the surface is:
$S=xy+2xz+2yz$
From the first equation we get:
$z=\frac{171.5}{xy} ; k=171.5\\ z=\frac{k}{xy}$
I will use k instead of a number just for the conveince.
We plug this into the second equation and we get:
$S=xy+2k\frac{1}{x}+2k\frac{1}{y}$
To find the minimum of this function we have to find the zeros of its first derivative.
Sx will denote the first derivative with respect to x and Sy will denote the first derivative with respect to Sy.
$S_x=y-2k\frac{1}{x^2}\\ S_y=x-2k\frac{1}{y^2}$
Now let both derivatives go to zero and solve the system (this will give us the so-called critical points).
$0=y-2k\frac{1}{x^2}\\ 0=x-2k\frac{1}{y^2}\\ y=2k\frac{1}{x^2}\\ x=2k\frac{1}{y^2}$
Now we plug in the first equation into the other and we get:
$x=\frac{\frac{2k}{1}}{\frac{4k^2}{x^4}}\\ x^3=2k\\ x=(2\cdot171.5)^{1/3}\\ x=7$
Now we can calculate y:
$y=2k\frac{1}{x^2}\\ y=2\cdot 171.5\frac{1}{7^2}=7$
And finaly we calculate z:
$z=\frac{171.5}{xy}\\ z=\frac{171.5}{7\cdot7}=3.5$
And finaly let's check our result:
$V=xyz=7\cdot7\cdot3.5=171.5$