Problem 19 - Calculus for Biology and Medicine

Question

Maximize the function $$f(x, y)=2 x y-x^{2} y-x y^{2}$$ on the triangle bounded by the line \(x+y=2\), the \(x\) -axis, and the \(y\) -axis.

Step-by-Step Solution

Answer

The maximum value of the function is 1.0 at \( (x, y) = (0.5, 1.5) \).

Step 1: Define the function and constraints

The function we want to maximize is \(f(x, y) = 2xy - x^2y - xy^2\). The region of interest is the triangle bounded by the lines \(x+y=2\), the \(x\)-axis \((y=0)\), and the \(y\)-axis \((x=0)\).

Step 2: Find derivatives

Compute the partial derivatives to find critical points. The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 2y - 2xy - y^2\). The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 2x - x^2 - 2xy\). Set both partial derivatives equal to zero to find critical points.

Step 3: Solve the system of equations

Set the partial derivatives to zero:1. \(2y - 2xy - y^2 = 0\). 2. \(2x - x^2 - 2xy = 0\).Solving these equations simultaneously can find critical points inside the region \(x \, \text{in} \, [0,2], \, y \, \text{in} \, [0,2]\).

Step 4: Check boundary conditions

Evaluate the function \(f(x, y)\) on the boundaries of the region:1. \(x = 0\): \(f(0, y) = 0\) for \(y = 0\to 2\)2. \(y = 0\): \(f(x, 0) = 0\) for \(x = 0\to 2\)3. On the line \(x+y=2\), substitute \(y=2-x\) into \(f(x, y)\) to get \(f(x, 2-x) = 2x(2-x) - x^2(2-x) - x(2-x)^2\).

Step 5: Optimize along the constraint \(x + y = 2\)

Substituting \(y = 2-x\) into \(f(x, y)\), we have:\[ f(x, 2-x) = 2x(2-x) - x^2(2-x) - x(2-x)^2 \]Which simplifies to \[ f(x, 2-x) = 4x - 4x^2 \]Find the derivative with respect to \(x\) and solve for critical points: \[ \frac{d}{dx}(4x - 4x^2) = 4 - 8x = 0 \]This yields the critical point \(x = 0.5\).

Step 6: Evaluate and compare results

Evaluate \(f(x,y)\) for the critical point and at the endpoints:1. At boundary \(x = 2, \, y = 0\) or \(x=0, \, y=2\): \(f(x, y) = 0\).2. At critical point \(x = 0.5, \, y = 1.5\) on the line \(x + y = 2\): \(f(0.5, 1.5) = 4(0.5) - 8(0.5)^2 = 1.0\). Thus, the maximum occurs at \((x, y) = (0.5, 1.5)\) with \(f(x, y) = 1.0\).

Key Concepts

Partial Derivatives

In calculus, partial derivatives represent the rate at which a function changes with respect to one of the variables, while all other variables are held constant.
For functions of multiple variables, like our function \(f(x, y) = 2xy - x^2y - xy^2\), partial derivatives help us explore how the function behaves by focusing on each variable independently.

• To find the partial derivative with respect to \(x\), we treat \(y\) as a constant. This gives us: \(\frac{\partial f}{\partial x} = 2y - 2xy - y^2\).
• For the partial derivative with respect to \(y\), treat \(x\) as constant: \(\frac{\partial f}{\partial y} = 2x - x^2 - 2xy\).

These derivatives are foundational in finding critical points, which are essential in optimization contexts.

Critical Points

Critical points occur where the partial derivatives of a function are zero, or where they do not exist. In the context of our optimization problem, finding these points is vital, as they potentially indicate where maximum or minimum values occur.
To find the critical points for our function, we set

• \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\).
• These translate into solving the system of equations: \(2y - 2xy - y^2 = 0\) and \(2x - x^2 - 2xy = 0\).

Solutions to these equations within our region of interest \((x, y)\) provide the critical points necessary for optimization.Additionally, we must examine these points with respect to the given constraints to determine feasibility and optimality.

Boundary Conditions

When optimizing a function over a defined region, examining its boundaries is crucial. The boundaries may themselves contain the maximum or minimum values of the function, especially in constrained domains like triangles, as in our problem.
The triangle in our problem is bounded by:

• The line \(x + y = 2\)
• The \(x\)-axis \((y = 0)\).
• The \(y\)-axis \((x = 0)\).

Evaluating the function \(f(x, y)\) along these boundaries helps uncover any potential extreme values.For instance:

• Along the line \(x + y = 2\), we substitute \(y = 2 - x\) and check the resulting function.
• For segments \(x = 0\) and \(y = 0\), setting one variable to zero simplifies the function evaluation to easily spot zero points.

Paying attention to boundary conditions ensures that no potential maximum or minimum point is overlooked.

Constraint Optimization

Constraint optimization entails finding maximum or minimum values of a function within a specific, often limited, region defined by constraints.
Many real-world problems, like our example, function under such constraints.The constraint \(x + y = 2\) establishes a clear boundary for our exploration.

• To maximize \(f(x, y)\) within this boundary, we simplify the function by substituting the constraint into the function: \(f(x, 2-x)\).
• Solving \(\frac{d}{dx}(4x - 4x^2) = 0\) yields the critical point \(x = 0.5\), which we subsequently validate.

Evaluating the function at this critical point confirms whether it is the highest or lowest value under the stated constraints.Constraint optimization allows mathematicians and practitioners to navigate real-life conditions, ensuring workable solutions.