If atmosphere pressure at a height of h units is given by the function `P(h)=he^(-h)` then pressure is maximum at the height of

Options

A. 1 unit
B. `(1)/(sqrt(e))` units
C. e units
D. 2 units

Answer

To find the height at which atmospheric pressure is maximum, we will follow these steps:

### Step 1: Define the function
The atmospheric pressure at a height \( h \) is given by the function:
\[ P(h) = h e^{-h} \]

### Step 2: Differentiate the function
To find the maximum pressure, we need to find the derivative of \( P(h) \) with respect to \( h \):
\[ P'(h) = \frac{d}{dh}(h e^{-h}) \]
Using the product rule, where \( u = h \) and \( v = e^{-h} \):
\[ P'(h) = u'v + uv' = (1)(e^{-h}) + (h)(-e^{-h}) = e^{-h} - h e^{-h} \]
Thus, we have:
\[ P'(h) = e^{-h}(1 - h) \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ e^{-h}(1 - h) = 0 \]
Since \( e^{-h} \) is never zero, we can focus on:
\[ 1 - h = 0 \implies h = 1 \]

### Step 4: Determine if it is a maximum
To confirm that this point is a maximum, we need to find the second derivative \( P''(h) \):
\[ P''(h) = \frac{d}{dh}(P'(h)) = \frac{d}{dh}(e^{-h}(1 - h)) \]
Using the product rule again:
\[ P''(h) = (-e^{-h})(1 - h) + e^{-h}(-1) = -e^{-h}(1 - h) - e^{-h} = -e^{-h}(2 - h) \]

Now, we evaluate the second derivative at \( h = 1 \):
\[ P''(1) = -e^{-1}(2 - 1) = -e^{-1} \]
Since \( -e^{-1} < 0 \), this indicates that \( P(h) \) has a local maximum at \( h = 1 \).

### Conclusion
The atmospheric pressure is maximum at a height of:
\[ \boxed{1} \]