Questions: The population density of a city is given by P(x, y)=-30 x^2-20 y^2+540 x+320 y+200, where x and y are miles from the southwest corner of the city limits and P is the number of people per square mile. Find the maximum population density, and specify where it occurs. The maximum density is people per square mile at (x, y)=

The population density of a city is given by P(x, y)=-30 x^2-20 y^2+540 x+320 y+200, where x and y are miles from the southwest corner of the city limits and P is the number of people per square mile. Find the maximum population density, and specify where it occurs.

The maximum density is people per square mile at (x, y)=

Transcript text: The population density of a city is given by $P(x, y)=-30 x^{2}-20 y^{2}+540 x+320 y+200$, where $x$ and $y$ are miles from the southwest corner of the city limits and P is the number of people per square mile. Find the maximum population density, and specify where it occurs. The maximum density is $\square$ people per square mile at $(x, y)=$ $\square$

Solution

Solution Steps

To find the maximum population density, we need to determine the critical points of the function $ P(x, y) $. This involves taking the partial derivatives of $ P $ with respect to $ x $ and $ y $, setting them to zero, and solving the resulting system of equations. Once we find the critical points, we can evaluate $ P $ at these points to find the maximum density.

Step 1: Identify the Function

The population density function is given by:

\[ P(x, y) = -30x^2 - 20y^2 + 540x + 320y + 200 \]

This is a quadratic function in two variables, which represents a paraboloid.

Step 2: Find the Critical Points

To find the maximum population density, we need to find the critical points by taking the partial derivatives and setting them to zero.

Partial Derivative with respect to $x$:

\[ \frac{\partial P}{\partial x} = -60x + 540 \]

Set $\frac{\partial P}{\partial x} = 0$:

\[ -60x + 540 = 0 \]

\[ x = \frac{540}{60} = 9 \]

Partial Derivative with respect to $y$:

\[ \frac{\partial P}{\partial y} = -40y + 320 \]

Set $\frac{\partial P}{\partial y} = 0$:

\[ -40y + 320 = 0 \]

\[ y = \frac{320}{40} = 8 \]

Step 3: Verify Maximum with Second Derivative Test

The second derivative test for functions of two variables involves the Hessian matrix:

\[ H = \begin{bmatrix} \frac{\partial^2 P}{\partial x^2} & \frac{\partial^2 P}{\partial x \partial y} \\ \frac{\partial^2 P}{\partial y \partial x} & \frac{\partial^2 P}{\partial y^2} \end{bmatrix} \]

Calculate the second derivatives:

\[ \frac{\partial^2 P}{\partial x^2} = -60 \]

\[ \frac{\partial^2 P}{\partial y^2} = -40 \]

\[ \frac{\partial^2 P}{\partial x \partial y} = 0 \]

The Hessian matrix is:

\[ H = \begin{bmatrix} -60 & 0 \\ 0 & -40 \end{bmatrix} \]

The determinant of $H$ is:

\[ \det(H) = (-60)(-40) - (0)(0) = 2400 \]

Since $\det(H) > 0$ and $\frac{\partial^2 P}{\partial x^2} < 0$, the critical point $(9, 8)$ is a local maximum.

Step 4: Calculate the Maximum Population Density

Substitute $x = 9$ and $y = 8$ into the original function:

\[ P(9, 8) = -30(9)^2 - 20(8)^2 + 540(9) + 320(8) + 200 \]