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miv72 [106K]
3 years ago
15

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.
Mathematics
1 answer:
____ [38]3 years ago
6 0

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

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