Question

A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 34.5 ft3. what dimensions yield the minimum surface area? round to the nearest tenth, if necessary.

Answer 1

Answer:

$l \approx 3.3\,ft$, $x = 3.2\,ft$

Step-by-step explanation:

The equations of volume and surface area are presented below:

$34.5 = l^{2}\cdot x$

$A_{s} = 2\cdot l^{2} + 4\cdot l \cdot x$

The length of the tank is:

$x = \frac{34.5}{l^{2}}$

The expression fo the surface area is therefore simplified into an univariable form:

$A_{s} = 2\cdot l^{2} + 4\cdot \left(\frac{34.5}{l} \right)$

$A_{s} = 2\cdot l^{2} + \frac{138}{l}$

The first and second derivatives of the expression are, respectively:

$A_{s}' = 4\cdot l -\frac{138}{l^{2}}$

$A_{s}'' = 4 +\frac{276}{l^{3}}$

The first derivative is equalized to zero and length of the square side is now found:

$4\cdot l -\frac{138}{l^{2}} = 0$

$4\cdot l^{3}-138 = 0$

$l^{3} = \frac{138}{4}$

$l = \sqrt[3]{\frac{138}{4} }$

$l \approx 3.3\,ft$

Now, the second derivative offers a criteria to determine if solution leads to an absolute minimum:

$A_{s}'' = 4 + \frac{276}{(3.3\,ft)^{3}}$

$A_{s}'' = 11.7$ (Absolute minimum)

The depth of the tank is:

$x = \frac{34.5\,ft^{3}}{(3.3\,ft)^{2}}$

$x = 3.2\,ft$