Find the four partial derivatives of P.

StartFraction partial derivative Upper P Over partial derivative w EndFraction∂P∂wequals=enter your response here

​(Simplify your answer. Use integers or decimals for any numbers in the expression. Round to six decimal places as​ needed.)

Part 3

StartFraction partial derivative Upper P Over partial derivative r EndFraction∂P∂requals=enter your response here

​(Simplify your answer. Use integers or decimals for any numbers in the expression. Round to six decimal places as​ needed.)

Part 4

StartFraction partial derivative Upper P Over partial derivative s EndFraction∂P∂sequals=enter your response here

​(Simplify your answer. Use integers or decimals for any numbers in the expression. Round to six decimal places as​ needed.)

Part 5

StartFraction partial derivative Upper P Over partial derivative t EndFraction∂P∂tequals=enter your response here

​(Simplify your answer. Use integers or decimals for any numbers in the expression. Round to six decimal places as​ needed.)

To find the four partial derivatives of \( P \) with respect to \( w, r, s, \) and \( t \), we will compute each one separately using the formula:

\[
P(w, r, s, t) = 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473}
\]

### 1. Partial Derivative of \( P \) with respect to \( w \)

\[
\frac{\partial P}{\partial w} = \frac{\partial}{\partial w} \left( 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473} \right)
\]

By applying the power rule, we have:

\[
\frac{\partial P}{\partial w} = 0.007868 \cdot (-0.645) \cdot w^{-0.645 - 1} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473}
\]

Simplifying:

\[
\frac{\partial P}{\partial w} = -0.00507286 \cdot w^{-1.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473}
\]

### 2. Partial Derivative of \( P \) with respect to \( r \)

\[
\frac{\partial P}{\partial r} = \frac{\partial}{\partial r} \left( 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473} \right)
\]

Applying the power rule:

\[
\frac{\partial P}{\partial r} = 0.007868 \cdot w^{-0.645} \cdot (1.039) \cdot r^{1.039 - 1} \cdot s^{0.873} \cdot t^{2.473}
\]

Simplifying:

\[
\frac{\partial P}{\partial r} = 0.008175652 \cdot w^{-0.645} \cdot r^{0.039} \cdot s^{0.873} \cdot t^{2.473}
\]

### 3. Partial Derivative of \( P \) with respect to \( s \)

\[
\frac{\partial P}{\partial s} = \frac{\partial}{\partial s} \left( 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473} \right)
\]

Applying the power rule:

\[
\frac{\partial P}{\partial s} = 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot (0.873) \cdot s^{0.873 - 1} \cdot t^{2.473}
\]

Simplifying:

\[
\frac{\partial P}{\partial s} = 0.006870564 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{-0.127} \cdot t^{2.473}
\]

### 4. Partial Derivative of \( P \) with respect to \( t \)

\[
\frac{\partial P}{\partial t} = \frac{\partial}{\partial t} \left( 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{2.473} \right)
\]

Applying the power rule:

\[
\frac{\partial P}{\partial t} = 0.007868 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot (2.473) \cdot t^{2.473 - 1}
\]

Simplifying:

\[
\frac{\partial P}{\partial t} = 0.019458364 \cdot w^{-0.645} \cdot r^{1.039} \cdot s^{0.873} \cdot t^{1.473}
\]

Now, let's calculate these derivatives with \( w = 21 \), \( r = 60 \), \( s = 420,000 \), and \( t = 8 \), rounding to six decimal places.The four partial derivatives of \( P \) are:

1. \(\frac{\partial P}{\partial w} = -33,136.805874\)
2. \(\frac{\partial P}{\partial r} = 18,682.479777\)
3. \(\frac{\partial P}{\partial s} = 2.242514\)
4. \(\frac{\partial P}{\partial t} = 333,506.538650\)

These values are rounded to six decimal places as requested.

Would you like any further details or have questions about these calculations?

Here are some related questions you might consider:
1. How does each variable impact profit sensitivity?
2. What happens if the occupancy rate \( r \) increases by 10%?
3. How would a change in reimbursement rate \( t \) affect profit?
4. What is the relationship between facility size \( s \) and annual profit?
5. How can this model be adjusted for different regions?

**Tip:** Partial derivatives help in understanding how a small change in one variable affects the overall outcome while keeping others constant.

Learn how to calculate the partial derivatives of function P with respect to variables w, r, s, and t. This problem involves applying the power rule in calculus to find derivatives, providing insights into sensitivity analysis in mathematical modeling.

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