Optimisation Problems

Identification of Local Maximum/Minimum Values

Definition

Local maximum: A point where the function’s value is greater than or equal to the values at all nearby points. It’s a ‘peak’ in the local area.

Local minimum: A point where the function’s value is less than or equal to the values at all nearby points. It’s a ‘valley’ in the local area.

Finding Local Extrema

1. Find the derivative: Calculate \(f'(x)\), the first derivative of the function \(f(x)\).
2. Find critical points: Solve \(f'(x) = 0\) to find critical points. These are potential locations of local maxima or minima.
3. Second derivative test (optional):
   • Calculate \(f''(x)\), the second derivative of the function.
   • If \(f''(x) > 0\) at a critical point, it’s a local minimum.
   • If \(f''(x) < 0\) at a critical point, it’s a local maximum.
   • If \(f''(x) = 0\), the test is inconclusive; use the first derivative test.
4. First derivative test:
   • Examine the sign of \(f'(x)\) to the left and right of each critical point.
   • If \(f'(x)\) changes from positive to negative, it’s a local maximum.
   • If \(f'(x)\) changes from negative to positive, it’s a local minimum.

Application to Solving Optimisation Problems

General Strategy

1. Understand the problem: Read the problem carefully and identify the quantity to be maximised or minimised (the objective function). Draw a diagram if possible.
2. Define variables: Assign variables to the relevant quantities.
3. Formulate the objective function: Express the objective function in terms of the defined variables. This will be \(y = f(x)\).
4. Identify constraints: Determine any constraints on the variables. These can often be expressed as equations or inequalities.
5. Reduce to a single variable: Use the constraints to eliminate variables and express the objective function in terms of a single independent variable. \(y = f(x)\)
6. Find critical points: Find the derivative of the objective function, \(f'(x)\), and solve for \(f'(x) = 0\) to find the critical points.
7. Check endpoints: If the function is defined on a closed interval \([a, b]\), evaluate the objective function at the endpoints \(f(a)\) and \(f(b)\).
8. Determine the maximum or minimum: Compare the values of the objective function at the critical points and endpoints to determine the absolute maximum or minimum.
9. Answer the question: State the solution in the context of the original problem.

Example

Find the maximum area of a rectangular field that can be enclosed by 100 m of fencing.

1. Objective: Maximise area \(A\).
2. Variables: Let the length be \(l\) and the width be \(w\).
3. Objective function: \(A = l \times w\)
4. Constraint: Perimeter is 100 m: \(2l + 2w = 100\), which simplifies to \(l + w = 50\).
5. Reduce to single variable: \(w = 50 - l\), so \(A = l(50 - l) = 50l - l^2\).
6. Find critical points: \(\frac{dA}{dl} = 50 - 2l\). Set \(\frac{dA}{dl} = 0\), so \(50 - 2l = 0\), which gives \(l = 25\).
7. Check endpoints (implicit): Since \(l\) and \(w\) must be positive, \(0 < l < 50\). As \(l\) approaches 0 or 50, the area approaches 0, which is a minimum.
8. Determine maximum: When \(l = 25\), \(w = 50 - 25 = 25\). The maximum area is \(A = 25 \times 25 = 625\) square meters.
9. Answer: The maximum area is 625 square meters when the field is a square with sides of 25 meters.

Identification of Interval Endpoint Maximum and Minimum Values

Importance

When optimising a function over a closed interval \([a, b]\), the absolute maximum or minimum may occur at the endpoints \(x = a\) or \(x = b\), rather than at a critical point within the interval.

Procedure

1. Find critical points: Identify all critical points of \(f(x)\) within the interval \([a, b]\).
2. Evaluate function at critical points: Calculate the function value \(f(x)\) at each critical point found in step 1.
3. Evaluate function at endpoints: Calculate the function values at the endpoints of the interval, \(f(a)\) and \(f(b)\).
4. Compare values: Compare the function values obtained in steps 2 and 3. The largest value is the absolute maximum, and the smallest value is the absolute minimum on the interval.

Example

Find the absolute maximum and minimum values of \(f(x) = x^3 - 6x^2 + 5\) on the interval \([-2, 5]\).

1. Find critical points:
   • \(f'(x) = 3x^2 - 12x\)
   • Set \(f'(x) = 0\): \(3x^2 - 12x = 0 \implies 3x(x - 4) = 0 \implies x = 0, x = 4\)
   • Both critical points, \(x = 0\) and \(x = 4\), lie within the interval \([-2, 5]\).
2. Evaluate at critical points:
   • \(f(0) = 0^3 - 6(0)^2 + 5 = 5\)
   • \(f(4) = 4^3 - 6(4)^2 + 5 = 64 - 96 + 5 = -27\)
3. Evaluate at endpoints:
   • \(f(-2) = (-2)^3 - 6(-2)^2 + 5 = -8 - 24 + 5 = -27\)
   • \(f(5) = (5)^3 - 6(5)^2 + 5 = 125 - 150 + 5 = -20\)
4. Compare values:
   • \(f(0) = 5\)
   • \(f(4) = -27\)
   • \(f(-2) = -27\)
   • \(f(5) = -20\)

Therefore, the absolute maximum value is 5 at \(x = 0\), and the absolute minimum value is -27 at \(x = -2\) and \(x = 4\).

Common Mistakes

• Forgetting to check endpoints when finding absolute extrema on a closed interval.
• Not verifying whether critical points are local maxima or minima using the first or second derivative test.
• Not properly defining the variables and objective function.
• Making algebraic errors when simplifying equations.