Problem 16 - Calculus for Biology and Medicine

Question

The functions are defined on the rectangular domain $$D=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\}$$ Find the global maxima and minima of \(f\) on \(D .\) $$ f(x, y)=x^{2}+y^{2} / 4 $$

Step-by-Step Solution

Answer

The global minimum is 0 at (0, 0), and the global maximum is \(\frac{5}{4}\) at the corners.

Step 1: Identify the domain and the function

We are given a function defined by \( f(x, y) = x^2 + \frac{y^2}{4} \) and a domain \( D = \{ (x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1 \} \). The function is quadratic, making it suitable for differentiating to find critical points.

Step 2: Find partial derivatives

Determine the partial derivatives of \( f(x, y) \). Let \( f_x \) and \( f_y \) denote the partial derivatives with respect to \( x \) and \( y \) respectively. Compute \( f_x = \frac{\partial}{\partial x} (x^2 + \frac{y^2}{4}) = 2x \) and \( f_y = \frac{\partial}{\partial y} (x^2 + \frac{y^2}{4}) = \frac{y}{2} \).

Step 3: Solve for critical points

Find where both partial derivatives are zero. Set \( f_x = 0 \) and \( f_y = 0 \), which gives the equations \( 2x = 0 \) and \( \frac{y}{2} = 0 \). Solving these yields \( x = 0 \) and \( y = 0 \). Thus, \((0, 0)\) is the critical point.

Step 4: Evaluate at critical point

Substitute the critical point \((0, 0)\) into the function \( f(x, y) = x^2 + \frac{y^2}{4} \). This gives \( f(0, 0) = 0^2 + \frac{0^2}{4} = 0 \).

Step 5: Examine the boundaries

Check the function values on the boundary of \( D \). The boundaries are given by \( x = 1, x = -1, y = 1, y = -1 \). Evaluate: - When \( x = 1 \): \( f(1, y) = 1^2 + \frac{y^2}{4} \).- When \( x = -1 \): \( f(-1, y) = (-1)^2 + \frac{y^2}{4} \).- When \( y = 1 \): \( f(x, 1) = x^2 + \frac{1^2}{4} \).- When \( y = -1 \): \( f(x, -1) = x^2 + \frac{(-1)^2}{4} \).For these linear boundaries, maximize and minimize within \(-1 \leq x, y \leq 1\).

Step 6: Calculate function values on corners

Evaluate \( f(x, y) \) at each corner of the domain \( D \):- \( f(-1, -1) = (-1)^2 + \frac{(-1)^2}{4} = 1 + \frac{1}{4} = \frac{5}{4} \).- \( f(-1, 1) = (-1)^2 + \frac{1^2}{4} = 1 + \frac{1}{4} = \frac{5}{4} \).- \( f(1, -1) = 1 + \frac{1}{4} = \frac{5}{4} \).- \( f(1, 1) = 1 + \frac{1}{4} = \frac{5}{4} \).

Step 7: Determine global extrema

Compare all calculated function values: critical point value is \( 0 \) and corner values are \( \frac{5}{4} \). The global minimum is at \((0, 0)\) with value \( 0 \) since it is the lowest. The global maximum is \( \frac{5}{4} \) at the corners, as they are the highest values in this case.

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of calculus to functions with more than one variable. While single-variable calculus deals with functions like \( f(x) \), multivariable calculus extends this concept to functions involving two or more variables, such as \( f(x, y) \). This branch of mathematics is crucial in analyzing real-world phenomena as it lets us handle complex situations where multiple factors are at play at once.
In our exercise, we deal with a function \( f(x, y) = x^2 + \frac{y^2}{4} \) defined over a rectangular domain. We aim to find where this function reaches its largest and smallest values, known as global extrema. Understanding multivariable calculus allows us to utilize techniques like taking partial derivatives to find critical points and analyzing boundary conditions to identify extrema. These are fundamental skills for solving problems involving multiple changing variables, particularly in fields like physics and engineering.

Critical Points

Critical points in multivariable calculus are similar to what they are in single-variable calculus, but they involve each variable within a function. For a function \( f(x,y) \), a critical point occurs where the partial derivatives \( f_x \) and \( f_y \) are both zero or undefined. This means the rate of change in any direction is zero, making it a potential candidate for points of interest such as minima, maxima, or saddle points.
In our exercise, we find the partial derivatives: \( f_x = 2x \) and \( f_y = \frac{y}{2} \). Setting these equal to zero reveals the critical point to be \((0,0)\). It's crucial to evaluate these points as they can indicate extrema within the domain. However, finding critical points isn't sufficient for determining global extrema – evaluating the function on the boundaries and corners of the domain is equally important to capture all potential extrema.

Rectangular Domain

The concept of a rectangular domain is straightforward – it's a defined area on the coordinate plane, bounded by constant values of \(x\) and \(y\). In our problem, the domain is defined as \( D = \{ (x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1 \} \). This means that the function \( f(x, y) \) is only considered within this range of \(x\) and \(y\).
It's essential to recognize the constraints of a rectangular domain, as they guide the evaluation of the function on the edges and corners beyond just the critical points. For example, evaluating the function on the boundary means checking lines where \( x = 1 \), \( y = 1 \), and so forth, while the corners like \((-1, -1)\) need to be examined directly. These practices ensure we've identified any possible extrema at the edges or corners of the domain, which sometimes can be the points where global maxima or minima actually occur.