Answer:
[tex]l \approx 3.3\,ft[/tex], [tex]x = 3.2\,ft[/tex]
Step-by-step explanation:
The equations of volume and surface area are presented below:
[tex]34.5 = l^{2}\cdot x[/tex]
[tex]A_{s} = 2\cdot l^{2} + 4\cdot l \cdot x[/tex]
The length of the tank is:
[tex]x = \frac{34.5}{l^{2}}[/tex]
The expression fo the surface area is therefore simplified into an univariable form:
[tex]A_{s} = 2\cdot l^{2} + 4\cdot \left(\frac{34.5}{l} \right)[/tex]
[tex]A_{s} = 2\cdot l^{2} + \frac{138}{l}[/tex]
The first and second derivatives of the expression are, respectively:
[tex]A_{s}' = 4\cdot l -\frac{138}{l^{2}}[/tex]
[tex]A_{s}'' = 4 +\frac{276}{l^{3}}[/tex]
The first derivative is equalized to zero and length of the square side is now found:
[tex]4\cdot l -\frac{138}{l^{2}} = 0[/tex]
[tex]4\cdot l^{3}-138 = 0[/tex]
[tex]l^{3} = \frac{138}{4}[/tex]
[tex]l = \sqrt[3]{\frac{138}{4} }[/tex]
[tex]l \approx 3.3\,ft[/tex]
Now, the second derivative offers a criteria to determine if solution leads to an absolute minimum:
[tex]A_{s}'' = 4 + \frac{276}{(3.3\,ft)^{3}}[/tex]
[tex]A_{s}'' = 11.7[/tex] (Absolute minimum)
The depth of the tank is:
[tex]x = \frac{34.5\,ft^{3}}{(3.3\,ft)^{2}}[/tex]
[tex]x = 3.2\,ft[/tex]
