Problem 18 - Calculus for Biology and Medicine

Question

Find the global 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

The global maxima is 0 at \((0,0)\) and \((-4,0)\), and the global minima is -3.75 at \((-2, \frac{1}{2})\).

Step 1: Identify Critical Points

To find the critical points of the function, we need to take the first partial derivatives and set them equal to zero. The partial derivative with respect to \(x\) is \(f_x = 2x + 4\), and with respect to \(y\) is \(f_y = -2y + 1\). Set these equal to zero: \(2x + 4 = 0\) gives \(x = -2\) and \(-2y + 1 = 0\) gives \(y = \frac{1}{2}\). We have a critical point at \((-2, \frac{1}{2})\).

Step 2: Verify Critical Point in Domain

Check if the critical point \((-2, \frac{1}{2})\) lies within the set \(D = \{-4 \leq x \leq 0, 0 \leq y \leq 1\}\). Since \(-4 \leq -2 \leq 0\) and \(0 \leq \frac{1}{2} \leq 1\), the critical point is within the domain \(D\).

Step 3: Evaluate Function at Critical Point

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

Step 4: Evaluate Function at Boundary Points

Evaluate the function at various points on the boundary of \(D\):- When \(x = -4\) and \(y = 0, 1\): \[f(-4, 0) = (-4)^2 - 0^2 + 4(-4) + 0 = 0\] \[f(-4, 1) = (-4)^2 - 1^2 + 4(-4) + 1 = -1\]- When \(x = 0\) and \(y = 0, 1\): \[f(0, 0) = 0^2 - 0^2 + 4(0) + 0 = 0\] \[f(0, 1) = 0^2 - 1^2 + 4(0) + 1 = 0\]. - When \(y = 0\) and \(x = -4, 0\), calculated above: both are 0.- When \(y = 1\) and \(x = -4, 0\), calculated above: results are -1 and 0.

Step 5: Compare Values to Determine Extrema

Compare all obtained function values: - \((-2, \frac{1}{2})\) gives \(-3.75\) - \((-4,0)\) gives \(0\)- \((-4,1)\) gives \(-1\)- \((0,0)\) gives \(0\)- \((0,1)\) gives \(0\).This comparison shows the global maximum is \(0\) and the global minimum is \(-3.75\).

Key Concepts

Global Maxima and Minima

When we talk about finding the global maxima and minima of a function, we're looking for the highest and lowest values the function can achieve within a given domain. To identify these extreme values, it's essential to consider both the interior of the domain and its boundary.

• The global maximum is the highest function value within the entire domain, whereas the global minimum is the lowest.
• Finding these involves evaluating the function at critical points as well as various boundary points.

In our exercise, we identified the function is defined on a rectangular domain. We discovered the extreme values by considering both critical points found within the domain and values at the boundaries of the domain. Both steps are crucial to ensure that potential maxima and minima aren't overlooked. Be thorough in evaluating the function at all critical and boundary points. Only then can we confidently identify true global maxima and minima within the domain of interest.

Partial Derivatives

Partial derivatives are a fundamental concept in calculus optimization, especially for functions of multiple variables. They help us understand how a function changes as each variable is varied independently, while holding other variables constant.

• For our function \(f(x, y) = x^2 - y^2 + 4x + y\), we computed the partial derivatives with respect to \(x\) and \(y\).
• The partial derivative with respect to \(x\), \(f_x = 2x + 4\), measures the rate of change of \(f\) as \(x\) changes, with \(y\) constant.
• Similarly, the partial derivative with respect to \(y\), \(f_y = -2y + 1\), represents the rate of change of \(f\) as \(y\) varies, holding \(x\) constant.

Setting these partial derivatives to zero helps locate critical points within the domain where the slope is flat. These points could potentially be local maxima or minima. It is also important to check these critical points to ensure they lie within the given domain, as seen in our solutions.

Boundary Points

In the context of finding global extrema, checking boundary points is vital. The reason is simple—extrema might not always be inside the domain; they could lie on the edge. Boundary points are defined by the limits of the domain for each variable. Evaluating the function at these points ensures you’ve considered all possibilities.

• For our problem, the domain was bounded by \(-4 \leq x \leq 0\) and \(0 \leq y \leq 1\).
• We evaluated the function at these boundary points like \((-4, 0)\), \((0, 1)\), and others.
• Each value was then compared to identify a global maximum or minimum.

Skipping this step can lead to incorrect conclusions about the global extrema, missing points where the function might achieve its absolute highest or lowest value.