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?
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]