Find the absolute maximum and minimum values of $f$ on the set $D$: $f(x,y) = 4x + 6y - x^2 - y^2 + 7$, $D = \{(x, y) \mid 0 \leq x \leq 4, \quad 0 \leq y \leq 5 \}$

• absolute maximum value:
• absolute minimum value:

Neetesh Kumar | November 30, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.7 Question Number 9
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Step-by-step solution:

Step 1: Understand the problem and the region

The function $f(x, y)$ is a quadratic function of both $x$ and $y$. Since we are working on a closed rectangular region $D$, the extreme values of the function (maximum and minimum) will either occur at the interior critical points (if any) or at the boundary points of the region.

The region $D$ is a rectangle defined by:

• $0 \leq x \leq 4$
• $0 \leq y \leq 5$

Step 2: Find the partial derivatives

We begin by finding the first partial derivatives of $f(x, y)$ with respect to $x$ and $y$ to locate any critical points within the interior of $D$.

• Partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = 4 - 2x$

• Partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = 6 - 2y$

Step 3: Solve for critical points

Set each partial derivative equal to zero to find the critical points:

• For $\frac{\partial f}{\partial x} = 4 - 2x = 0$, we solve for $x$: $4 - 2x = 0 \quad \Rightarrow \quad x = 2$

• For $\frac{\partial f}{\partial y} = 6 - 2y = 0$, we solve for $y$: $6 - 2y = 0 \quad \Rightarrow \quad y = 3$

Thus, there is a critical point at $(x, y) = (2, 3)$. We now need to check if this critical point lies within the region $D$. Since $0 \leq x \leq 4$ and $0 \leq y \leq 5$, the point $(2, 3)$ is indeed inside the region $D$.

Step 4: Evaluate $f(x, y)$ at the critical point and boundary points

1. Evaluate at the interior critical point $(2, 3)$:

Substitute $x = 2$ and $y = 3$ into $f(x, y)$:

$f(2, 3) = 4(2) + 6(3) - (2)^2 - (3)^2 + 7$

$f(2, 3) = 8 + 18 - 4 - 9 + 7 = 20$

2. Evaluate at the boundary points:

We now evaluate $f(x, y)$ at the vertices of the rectangle $D$ (the boundary points) and along the edges.

Vertex $(0, 0)$:

$f(0, 0) = 4(0) + 6(0) - (0)^2 - (0)^2 + 7 = 7$

Vertex $(4, 0)$:

$f(4, 0) = 4(4) + 6(0) - (4)^2 - (0)^2 + 7 = 16 - 16 + 7 = 7$

Vertex $(0, 5)$:

$f(0, 5) = 4(0) + 6(5) - (0)^2 - (5)^2 + 7 = 30 - 25 + 7 = 12$

Vertex $(4, 5)$:

$f(4, 5) = 4(4) + 6(5) - (4)^2 - (5)^2 + 7 = 16 + 30 - 16 - 25 + 7 = 12$

Along the edge $x = 0$ (from $(0, 0)$ to $(0, 5)$):

Substitute $x = 0$ into $f(x, y)$:

$f(0, y) = 4(0) + 6y - (0)^2 - y^2 + 7 = 6y - y^2 + 7$

Evaluate at the endpoints:

• At $(0, 0)$, $f(0, 0) = 7$ (already calculated).
• At $(0, 5)$, $f(0, 5) = 12$ (already calculated).

This is a downward-opening quadratic function in $y$. Since $f(0, y) = 6y - y^2 + 7$ has a vertex at $y = 3$, the maximum value on this edge is $f(0, 3)$.

$f(0, 3) = 6(3) - (3)^2 + 7 = 18 - 9 + 7 = 16$

Along the edge $x = 4$ (from $(4, 0)$ to $(4, 5)$):

Substitute $x = 4$ into $f(x, y)$:

$f(4, y) = 4(4) + 6y - (4)^2 - y^2 + 7 = 16 + 6y - 16 - y^2 + 7 = 6y - y^2 + 7$

This is the same quadratic expression as along the edge $x = 0$, so we already know that the maximum value on this edge is $f(4, 3) = 16$. The values at the endpoints are:

• At $(4, 0)$, $f(4, 0) = 7$ (already calculated).
• At $(4, 5)$, $f(4, 5) = 12$ (already calculated).

Along the edge $y = 0$ (from $(0, 0)$ to $(4, 0)$):

Substitute $y = 0$ into $f(x, y)$:

$f(x, 0) = 4x + 6(0) - x^2 - (0)^2 + 7 = 4x - x^2 + 7$

This is a downward-opening quadratic function in $x$. We already know the values at the endpoints:

• At $(0, 0)$, $f(0, 0) = 7$ (already calculated).
• At $(4, 0)$, $f(4, 0) = 7$ (already calculated).

The maximum value along this edge occurs at $x = 2$, and we already know that $f(2, 0) = 7$.

Along the edge $y = 5$ (from $(0, 5)$ to $(4, 5)$):

Substitute $y = 5$ into $f(x, y)$:

$f(x, 5) = 4x + 6(5) - x^2 - (5)^2 + 7 = 4x + 30 - x^2 - 25 + 7 = -x^2 + 4x + 12$

This is a downward-opening quadratic function in $x$. We already know the values at the endpoints:

• At $(0, 5)$, $f(0, 5) = 12$ (already calculated).
• At $(4, 5)$, $f(4, 5) = 12$ (already calculated).

The maximum value along this edge occurs at $x = 2$, and we already know that $f(2, 5) = 16$.

Step 5: Conclusion

We have evaluated $f(x, y)$ at the vertices, along the edges, and at the critical point. The maximum value occurs at the point $(2, 3)$, and the minimum value occurs at the point $(4, 0)$ and $(0, 0)$.

• Absolute Maximum Value: $20$ at $(2, 3)$
• Absolute Minimum Value: $7$ at $(0, 0)$ and $(4, 0)$

Final Answer:

Absolute maximum value: $\boxed{20}$

Absolute minimum value: $\boxed{7}$

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