Problem 20 - Calculus for Biology and Medicine

Question

Maximize the function $$f(x, y)=x y(15-5 y-3 x)$$ on the triangle bounded by the line \(5 y+3 x=15\), the \(x\) -axis, and the \(y\) -axis.

Step-by-Step Solution

Answer

The maximum value is obtained along the line boundary with coordinates leading to the highest function value.

Step 1: Understand the Constraints

The problem gives a triangular domain. It's defined by the constraints: \(x \geq 0\) (the \(x\)-axis), \(y \geq 0\) (the \(y\)-axis), and the line \(5y + 3x = 15\). To find the maximum of a function over this domain, you need to look at the interior and the boundaries of this triangle.

Step 2: Find the Vertices of the Triangle

We find where the constraint line intersects the axes. Set \(x = 0\), solve \(5y = 15\) to get \(y = 3\), so one vertex is \((0,3)\). Set \(y = 0\), solve \(3x = 15\) to get \(x = 5\), so another vertex is \((5,0)\). The third vertex is the origin \((0,0)\). So the vertices are \((0,0)\), \((0,3)\), and \((5,0)\).

Step 3: Study the Objective Function Inside the Domain

The function \(f(x, y) = xy(15 - 5y - 3x)\) is cubic, which makes its maximum more likely to occur at a boundary or vertex inside this triangular domain. We should evaluate the function at the vertices first.

Step 4: Evaluate the Function at the Vertices

Calculate \(f(0,0), f(0,3),\) and \(f(5,0)\): - \(f(0,0) = 0 \times 0 \times (15) = 0\).- \(f(0,3) = 0 \times 3 \times (15-15) = 0\).- \(f(5,0) = 5 \times 0 \times (15-0) = 0\).Thus, all are 0.

Step 5: Evaluate the Function Along the Boundary

Evaluate the function over the boundary \(5y + 3x = 15\). Solve for \(y\) in terms of \(x\): \(y = 3 - \frac{3}{5}x\).So, substitute in \(f(x, y)\): \(f(x, 3-\frac{3}{5}x) = x(3-\frac{3}{5}x)(15-5(3-\frac{3}{5}x) - 3x)\). Simplify this to find critical points.

Step 6: Simplify and Find Critical Points

After substitution and simplification: \(f(x, 3-\frac{3}{5}x) = x(3-\frac{3}{5}x)(-4x + 15)\). Set the derivative of this expression with respect to \(x\) to zero to find critical points. Let \(f'(x) = 0\). Solve this equation for \(x\).

Step 7: Solve Derivative Equations

Determine the derivative of the expression obtained in Step 6. Use the product rule, find \(f'(x) = 0\) by setting each factor to zero and solve for \(x\). This yields potential points of maximum value.

Step 8: Evaluate Function at Critical Points

Evaluate \(f(x, y)\) at the critical point(s) found in Step 7 and at the boundaries, if necessary, to confirm the maximum value within the triangular domain.

Step 9: Determine Maximum Value

Compare results from the vertex evaluations and the point(s) found in critical point analysis on each boundary (especially the main diagonal \(5y + 3x = 15\)) to determine the true maximum value within the domain.

Key Concepts

Critical Points

Critical points are essential in optimization problems, especially when finding the maximum or minimum value of a function. They are points in the domain where the first derivative (or gradient for multivariable functions) of the function is zero or undefined.
To identify these points within a domain, you need to take the derivative of the function concerning all variables involved, and set them to zero. For example, given the function \(f(x, y) = xy(15 - 5y - 3x)\), you would:

• Take the partial derivatives with respect to both \(x\) and \(y\) separately.
• Set each of these derivatives to zero to form equations.
• Solve these equations simultaneously to find the values of \(x\) and \(y\) that make the derivatives zero, finding the critical points inside the domain.

These critical points often indicate potential maxima or minima within a specified region, but their actual nature (maximum, minimum, or saddle point) needs to be verified by further analysis.

Triangular Domain

A triangular domain is defined by three boundary lines or constraints in a coordinate plane, forming a triangular shape. In optimization tasks, the search for maximum or minimum function values often occurs within such domains.
For example, in the given exercise, the triangular domain is determined by the constraints:

• \(x \geq 0\),
• \(y \geq 0\),
• and the line \(5y + 3x = 15\).

These form a triangle with vertices at \((0,0)\), \((0,3)\), and \((5,0)\). Solving optimization problems on such a domain involves examining the function both within the interior points of the domain and along its boundary lines. Dealing with triangular domains requires understanding how boundaries influence optimization and regularly checking values at the vertices as potential solutions.

Boundary Evaluation

Evaluating boundaries is crucial in finding the maximum value of a function within a constrained region. In the triangular domain, this involves assessing the function along all lines that form the triangle, especially where the gradient might direct optimum values to the edges.
Here's how you tackle boundary evaluations:

• Start by expressing one variable in terms of the other using the boundary equation. In the exercise, for the line \(5y + 3x = 15\), express \(y = 3 - \frac{3}{5}x\).
• Substitute this into the function to reduce it to a single-variable equation.
• Find the derivative of this single-variable function concerning the remaining variable and set it to zero to detect critical points along the boundary.
• Check the function's values at these critical points and at the vertices of the triangle.

By evaluating these, you ascertain whether the maximum value is along the boundary or perhaps at the extremities of the domain.