Question

The area, in square feet, of the horizontal cross section at height h feet is modeled by the function f given by f(h)=(50.3)/(e⁽⁰.²ʰ⁾+h). Based on this model, find the volume of the tank. Indicate units of measure.

Answer 1

The height at which the tank has a maximum cross-sectional area is approximately
`\( (\ln(5))/(0.2) \)` feet. The corresponding maximum area is approximately 3.05 square feet.

Sure, let's go through the calculations step by step.

1. Find the Derivative \(f'(h)\):

`\[f'(h) = (50.3(-0.2e^(0.2h) + 1))/((e^(0.2h) + h)^2)\]`

2. Set \(f'(h)\) Equal to Zero:

`\[-0.2e^(0.2h) + 1 = 0\]`

`\[e^(0.2h) = (1)/(0.2)\]`

`\[e^(0.2h) = 5\]`

`\[0.2h = \ln(5)\]`

`\[h = (\ln(5))/(0.2)\]`

3. Calculate Maximum Cross-Sectional Area:

\[A_{\text{max}} = f\left(\frac{\ln(5)}{0.2}\right) = \frac{50.3}{e^{0.2\left(\frac{\ln(5)}{0.2}\right)} + \frac{\ln(5)}{0.2}}\]

Now, let's compute this value.

\[A_{\text{max}} = \frac{50.3}{e^{0.2 \cdot \frac{\ln(5)}{0.2}} + \frac{\ln(5)}{0.2}}\]

\[A_{\text{max}} = \frac{50.3}{e^{\ln(5)} + \frac{\ln(5)}{0
`\[A_{\text{max}} = f\left((\ln(5))/(0.2)\right) = \frac{50.3}{e^{0.2\left((\ln(5))/(0.2)\right)} + (\ln(5))/(0.2)}\]`.2}}\]

`\[A_{\text{max}} = (50.3)/(5 + (\ln(5))/(0.2))\]`

`\[A_{\text{max}} \approx (50.3)/(5 + 2.3026/0.2)\]`

`\[A_{\text{max}} \approx (50.3)/(5 + 11.513)\]`

`\[A_{\text{max}} \approx (50.3)/(16.513)\]`

`\[A_{\text{max}} \approx 3.05 \, \text{square feet}\]`

4. Calculate the Volume of the Tank:

`\[V = \int_(0)^{(\ln(5))/(0.2)} f(h) \, dh\]`

`\[V = \int_(0)^{(\ln(5))/(0.2)} (50.3)/(e^(0.2h) + h) \, dh\]`

The probable question maybe:

"At what height \( h \) does the tank have a maximum cross-sectional area, and what is the corresponding maximum cross-sectional area? Additionally, calculate the volume of the tank at this height. Please provide the units of measure for both the maximum cross-sectional area and the volume."

This question is asking for the critical point where the cross-sectional area is maximized, and then it proceeds to find the corresponding maximum area and the volume of the tank at that specific height.