Problem 18 - Calculus for Biology and Medicine

Question

Find the absolute maxima and minima of $$ f(x, y)=x^{2}-y^{2}+4 x+y $$ on the set $$ D=\\{(x, y)=-4 \leq x \leq 0,0 \leq y \leq 1\\} $$

Step-by-Step Solution

Answer

Max value is \(\frac{1}{4}\) and min value is \(-4\).

Step 1: Understanding the Function and Domain

We are given the function \( f(x, y) = x^2 - y^2 + 4x + y \) and need to find its absolute maxima and minima within the domain \( D = \{(x, y) \,|\, -4 \leq x \leq 0, \ 0 \leq y \leq 1\} \). This domain forms a rectangle in the xy-plane.

Step 2: Determine Critical Points

To find the critical points, we need the partial derivatives \( f_x \) and \( f_y \). Calculate:- \( f_x = \frac{\partial f}{\partial x} = 2x + 4 \)- \( f_y = \frac{\partial f}{\partial y} = -2y + 1 \)Set these equal to zero: \( 2x + 4 = 0 \) and \( -2y + 1 = 0 \).Solving these gives us:- \( x = -2 \)- \( y = \frac{1}{2} \)A critical point inside the domain is \((-2, \frac{1}{2})\).

Step 3: Evaluate the Function at Critical Points

Substitute the critical point \((-2, \frac{1}{2})\) into \( f(x, y) \):\[f(-2, \frac{1}{2}) = (-2)^2 - (\frac{1}{2})^2 + 4(-2) + \frac{1}{2} = 4 - \frac{1}{4} - 8 + \frac{1}{2} = -3.75\]

Step 4: Check Boundary of the Domain

Evaluate \( f(x, y) \) on each boundary line of the domain:- For \( x = -4 \): \( f(-4, y) = (-4)^2 - y^2 + 4(-4) + y = 16 - y^2 - 16 + y = -y^2 + y \) Max at \( y = \frac{1}{2} \), \( f(-4, \frac{1}{2}) = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4} \).- For \( x = 0 \): \( f(0, y) = 0^2 - y^2 + 4(0) + y = -y^2 + y \) Max at \( y = \frac{1}{2} \), \( f(0, \frac{1}{2}) = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4} \).- For \( y = 0 \): \( f(x, 0) = x^2 + 4x \) Critical at \( x = -2 \), \( f(-2, 0) = 4 - 8 = -4 \).- For \( y = 1 \): \( f(x, 1) = x^2 - 1 + 4x + 1 = x^2 + 4x \) Critical at \( x = -2 \), \( f(-2, 1) = -4 \).

Step 5: Compare Function Values

Critical and boundary evaluations yielded:- \( f(-2, \frac{1}{2}) = -3.75 \)- \( f(-4, \frac{1}{2}) = \frac{1}{4} \)- \( f(0, \frac{1}{2}) = \frac{1}{4} \)- \( f(-2, 0) = -4 \)- \( f(-2, 1) = -4 \)- Primarily at endpoints: \((-4, 0)\), \((0, 0)\), \((-4, 1)\), \((0, 1)\)The absolute maximum value is \( \frac{1}{4} \) and the absolute minimum value is \( -4 \).

Key Concepts

Understanding Critical Points

To comprehend absolute maxima and minima, it's essential to recognize critical points first. Critical points occur where the function's slope is zero or undefined. In the context of functions with two variables, these are identified by calculating partial derivatives and setting them to zero.

For the function \( f(x, y) = x^2 - y^2 + 4x + y \), we first determine the partial derivatives:

• The derivative with respect to \( x \), denoted as \( f_x \), is \( 2x + 4 \).
• The derivative with respect to \( y \), denoted as \( f_y \), is \( -2y + 1 \).

Setting these to zero to find critical points, we solve the equations:

- \( 2x + 4 = 0 \), giving \( x = -2 \)
- \( -2y + 1 = 0 \), giving \( y = \frac{1}{2} \)

So, the critical point is \((-2, \frac{1}{2})\). Validating if this point lies within our domain is key before considering it further in evaluating extrema.

Partial Derivatives and Their Evaluation

Partial derivatives allow us to understand how a function changes with respect to each variable independently. They are vital in finding the slope of the function and identifying points of horizontal tangency where potential extrema may occur.

For a function of two variables like our example, partial derivatives are calculated separately for each variable.

• \( f_x = \frac{\partial f}{\partial x} = 2x + 4 \)
• \( f_y = \frac{\partial f}{\partial y} = -2y + 1 \)

These results tell us:

- \( f_x = 0 \) implies how the function behaves purely along the x-axis while treating y as constant.
- \( f_y = 0 \) similarly implies behavior along the y-axis independent of x.

Together, these derivatives help pin down critical points and areas of interest when assessing the function’s extremities.

Evaluating Function on Boundaries

Boundary evaluation involves analyzing a function along the edges of the defined domain. It ensures that we do not miss potential extremal values that might occur not just at critical points, but also at boundaries or corners.

In our exercise, the boundaries of the domain \( D \) are:

• \( x = -4 \) to \( x = 0 \) with \( y \) varying from 0 to 1.
• \( y = 0 \) to \( y = 1 \) while \( x \) varies between -4 and 0.

Evaluating the function along these boundaries includes setting each boundary condition in the function and looking for maxima or minima:

- For \( x = -4 \) and \( x = 0 \), simplify to \(-y^2 + y\) and find an extremum at \( y = \frac{1}{2} \), resulting in a function value of \( \frac{1}{4} \).
- For boundaries \( y = 0 \) and \( y = 1 \), evaluate \( f(x, 0) \) and \( f(x, 1) \) to retrieve corresponding minimum values \( -4 \).

This thorough analysis ensures that all potential extremal values within the closed region are considered.