Problem 32 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$\begin{array}{l}{f(x, y)=4 x+6 y-x^{2}-y^{2}} \\ {D=\\{(x, y) | 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant 5\\}}\end{array}$$

Step-by-Step Solution

Answer

Max is 29 at (2, 5) and min is 0 at (0, 0).

Step 1: Find Critical Points Inside D

To find the critical points of the function \( f(x, y) = 4x + 6y - x^2 - y^2 \), first compute the partial derivatives.\[\frac{\partial f}{\partial x} = 4 - 2x \] \[\frac{\partial f}{\partial y} = 6 - 2y \]Set these equal to zero to find critical points:\[4 - 2x = 0 \implies x = 2 \] \[6 - 2y = 0 \implies y = 3 \]The critical point is \((2, 3)\). Check if it is within the domain \( D \). It is, as \(0 \leq 2 \leq 4\) and \(0 \leq 3 \leq 5\).

Step 2: Evaluate f at Critical Points

Evaluate \( f \) at the critical point \((2, 3)\):\[ f(2, 3) = 4(2) + 6(3) - 2^2 - 3^2 \]\[ = 8 + 18 - 4 - 9 = 13 \]

Step 3: Evaluate f on Boundary of D

Now evaluate \( f \) along the boundaries of \( D \).1. **Boundary 1:** At \( x = 0 \), \( y \) ranges from 0 to 5: \( f(0, y) = 6y - y^2 \) Differentiate and find critical points: \( \frac{d}{dy} (6y - y^2) = 6 - 2y = 0 \implies y = 3 \) Evaluate endpoints and critical point: \( f(0, 0) = 0 \), \( f(0, 3) = 9 \), \( f(0, 5) = 5 \)2. **Boundary 2:** At \( x = 4 \), \( y \) ranges from 0 to 5: \( f(4, y) = 16 + 6y - y^2 \) Differentiate: \( \frac{d}{dy} (16 + 6y - y^2) = 6 - 2y = 0 \implies y = 3 \) Evaluate endpoints and critical point: \( f(4, 0) = 16 \), \( f(4, 3) = 25 \), \( f(4, 5) = 21 \)3. **Boundary 3:** At \( y = 0 \), \( x \) ranges from 0 to 4: \( f(x, 0) = 4x - x^2 \) Differentiate: \( \frac{d}{dx} (4x - x^2) = 4 - 2x = 0 \implies x = 2 \) Evaluate endpoints and critical point: \( f(0, 0) = 0 \), \( f(2, 0) = 4 \), \( f(4, 0) = 0 \)4. **Boundary 4:** At \( y = 5 \), \( x \) ranges from 0 to 4: \( f(x, 5) = 30 + 4x - x^2 \) Differentiate: \( \frac{d}{dx} (30 + 4x - x^2) = 4 - 2x = 0 \implies x = 2 \) Evaluate endpoints and critical point: \( f(0, 5) = 25 \), \( f(2, 5) = 29 \), \( f(4, 5) = 21 \)

Step 4: Find the Absolute Maximum and Minimum

Now, compare all values computed from the critical point and boundaries:Critical point: \( f(2, 3) = 13 \).Boundary evaluations:- \( f(0, 0) = 0 \)- \( f(0, 3) = 9 \)- \( f(0, 5) = 5 \)- \( f(4, 0) = 16 \)- \( f(4, 3) = 25 \)- \( f(4, 5) = 21 \)- \( f(2, 0) = 4 \)- \( f(0, 5) = 25 \)- \( f(2, 5) = 29 \)The absolute maximum is \( f(2, 5) = 29 \) and the absolute minimum is \( f(0, 0) = 0 \).

Key Concepts

Critical Points

In multivariable calculus, a critical point is where the partial derivatives of a function are zero or undefined. This indicates that the function does not have a slope, and is flat in all directions at that point. To find the critical points of a function like \( f(x, y) \), we compute its partial derivatives with respect to each variable. For \( f(x, y) = 4x + 6y - x^2 - y^2 \), we find:

• Partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 4 - 2x \)
• Partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = 6 - 2y \)

Setting the partial derivatives equal to zero, we solve:

• \(4 - 2x = 0 \Rightarrow x = 2\)
• \(6 - 2y = 0 \Rightarrow y = 3\)

Giving a critical point of \((2, 3)\). This helps identify potential local extrema within the domain defined.

Boundary Evaluation

Evaluating boundaries is crucial to find global maximum and minimum values, especially when a function is defined over a closed and bounded domain. It involves calculating the function's value along the limits, or edges, of the domain. These boundaries in our problem are determined by \(0 \leq x \leq 4\) and \(0 \leq y \leq 5\).We must evaluate the function at these edges to ensure we have explored all possibilities. For example:

• At \( x = 0 \), we examine \( y \'s \) range from 0 to 5.
• Similarly, at \( x = 4 \), and \( y = 0 \) or \( y = 5 \).

By differentiating the functions \( f(0, y) \), \( f(4, y) \), \( f(x, 0) \), and \( f(x, 5) \), and setting these derivatives to zero, we identify critical points along each boundary and calculate the function values at both endpoints and critical points.

Partial Derivatives

Partial derivatives are fundamental in identifying changes in multivariable functions. They represent the rate of change of a function with respect to one variable while keeping others constant.For example, in \( f(x, y) = 4x + 6y - x^2 - y^2 \), the partial derivative with respect to \( x, \; \frac{\partial f}{\partial x} = 4 - 2x \), tells us how \( f \) changes as \( x \) changes when \( y \) is fixed. Similarly, \( \frac{\partial f}{\partial y} = 6 - 2y \) informs us about changes in \( f \) as \( y \) varies.These derivatives help in locating critical points where potential maxima or minima might occur. This was used to find that the function levels off at point \((2, 3)\) inside the domain.

Absolute Maximum and Minimum

The absolute maximum and minimum values of a function on a given domain can be determined by comparing the function values at critical points and along the boundaries.After finding critical points within the domain, it is necessary to evaluate the function at these points and along the edges (as in boundary evaluation). Then, we compare all these values:

• The computed value at the critical point \((2, 3)\) gives us \( f(2, 3) = 13 \).
• Boundary evaluations provide numerous other values such as \( f(4, 3) = 25 \), and \( f(2, 5) = 29 \).
• Finally, comparing all calculated function values helps us conclude that the absolute maximum is \( 29 \) at \((2, 5)\), and the absolute minimum is \( 0 \) at \((0, 0)\).

By thoroughly analyzing both critical points and boundaries, we ensure that we find the global extremum within the defined domain.