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