Problem 4 - Active Calculus Multivariable

Question

Let \(f(x, y)=1 / x+2 / y+3 x+4 y\) in the region \(R\) where \(x, y>0\). Explain why \(f\) must have a global minimum at some point in \(R\) (note that \(R\) is unbounded- - how does this influence your explanation?). Then find the global minimum. minimum \(=\) ______________.

Step-by-Step Solution

Answer

Minimum \(= f(\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{2}}) = \frac{1}{\sqrt{\frac{1}{3}}} + \frac{2}{\sqrt{\frac{1}{2}}} + 3\sqrt{\frac{1}{3}} + 4\sqrt{\frac{1}{2}} \)

Step 1: 1. Compute the Partial Derivatives

First, we need to compute the two first-order partial derivatives with respect to x and y. For x: \( \frac{\partial f}{\partial x} = -\frac{1}{x^2} + 3 \) For y: \( \frac{\partial f}{\partial y} = - \frac{2}{y^2} + 4 \)

Step 2: 2. Set the Partial Derivatives to Zero

In order to find the critical points, we should set the partial derivatives equal to zero and solve the system of equations. For x derivative: \( -\frac{1}{x^2} + 3 = 0 \) For y derivative: \( -\frac{2}{y^2} + 4 = 0 \)

Step 3: 3. Solve the System of Equations

Now, we will solve the system of equations to find the critical points: For x: \( x^2 = \frac{1}{3} \) \( x = \sqrt{\frac{1}{3}} \) For y: \( y^2 = \frac{1}{2} \) \( y = \sqrt{\frac{1}{2}} \) So, the critical point is \((\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{2}})\).

Step 4: 4. Determine if the Critical Point is a Minimum

Now we need to examine whether the critical point we found is a minimum. Since the given function is monotonic, it gets bigger as x and y get bigger, which means that the critical point we found is indeed a relatively minimum point in the region \(R\). For the given function, this relative minimum would be the global minimum, because the function increases as x and y values increase in the positive direction.

Step 5: 5. Calculate the Global Minimum

To find the global minimum value, we substitute the critical point into the original function: \( f(\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{2}}) = \frac{1}{\sqrt{\frac{1}{3}}} + \frac{2}{\sqrt{\frac{1}{2}}} + 3\sqrt{\frac{1}{3}} + 4\sqrt{\frac{1}{2}} \) Minimum \(= f(\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{2}}) \)

Key Concepts

Partial Derivatives

In calculus, partial derivatives are used to find the rate of change of a function with respect to one of its variables while keeping the other variables constant. For a function like \(f(x, y) = \frac{1}{x} + \frac{2}{y} + 3x + 4y\), we have two variables, \(x\) and \(y\).

To find the partial derivatives, we differentiate the function with respect to each variable individually:

• For the partial derivative with respect to \(x\), we treat \(y\) as a constant, resulting in \(\frac{\partial f}{\partial x} = -\frac{1}{x^2} + 3\).
• Similarly, for the partial derivative with respect to \(y\), treating \(x\) as a constant gives us \(\frac{\partial f}{\partial y} = -\frac{2}{y^2} + 4\).

Understanding and computing partial derivatives are crucial steps in finding where a function reaches its minimum or maximum values, especially when dealing with multiple variables.

Critical Points

Critical points are specific points on the graph of a function where the first derivatives (the slopes) are zero or undefined. These points are important as they help determine where functions might have local minima, maxima, or saddle points.

To find critical points, we set the partial derivatives equal to zero and solve for the variables. For our function \(f(x, y)\), we set:

• \(-\frac{1}{x^2} + 3 = 0\) to find critical \(x\)
• \(-\frac{2}{y^2} + 4 = 0\) to find critical \(y\)

Solving these gives the critical point \((\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{2}})\). Recognizing and solving for critical points are essential for identifying where a function may reach its peak or dip within a given region.

System of Equations

A system of equations is a set of two or more equations that share common variables. In solving such systems, we look for values that satisfy all given equations simultaneously. For functions depending on multiple variables, like in calculus problems, solving systems of equations often reveals critical points.

In our example, we have two equations obtained from the partial derivatives:

• \(-\frac{1}{x^2} + 3 = 0\)
• \(-\frac{2}{y^2} + 4 = 0\)

By solving this system, we determine points like \(x = \sqrt{\frac{1}{3}}\) and \(y = \sqrt{\frac{1}{2}}\). These solutions give us a deeper insight into the behavior of the function, guiding us to where it potentially achieves global minima or maxima.

Unbounded Region

An unbounded region is an area in the coordinate plane that extends infinitely in one or more directions. This concept is critical when analyzing where a function might achieve a global minimum or maximum because it impacts our understanding of the function's behavior at the region's edges.

In our problem's context, the region \(R\) is defined by \(x, y > 0\), meaning it's open and extends indefinitely in the positive \(x\) and \(y\) directions. This unbounded nature means that a function might theoretically continue to decrease or increase without bound.

However, the specific behavior of the function \(f(x, y)\) ensures that as \(x\) and \(y\) get very large, \(f\) also increases. This implies that there is a lowest value within the region, which corresponds to the global minimum. Identifying how unbounded regions affect the function aids in predicting and comprehending the optimization points, such as global minima or maxima.