If we optimize the function f(x)=4x2 - 24x + 132, the optimum will be at x = 3 , which is a local minimum .

Question

BlackTom AI · Accepted Answer

The question asks about optimizing the function f(x) = 4x^2 - 24x + 132.

First, identify the form: it is a quadratic with a = 4 > 0, so the parabola opens upward and has a single vertex which is a minimum point.

To locate the vertex (and thus the optimum), compute x = -b/(2a) where the quadratic is written as f(x) = ax^2 + bx + c. Here, b = -24 and a = 4, so x = -(-24) / (2 * 4) = 24 / 8 = 3.

Option analysis:

- For the x-coordinate: The candidate optimum is x = 3. This matches the vertex calculation above, so this option is correct.

- For the type of optimum: Since a > 0, the vertex represents a global (and local) minimum for a quadratic, not a maximum. Therefore, the statement that the optimum is a minimum is correct; the parabola’s lowest point is at x = 3.

If you were to state the minimum value explicitly, you would substitute x = 3 back into f(x): f(3) = 4(3)^2 - 24(3) + 132 = 36 - 72 + 132 = 96. So the minimum value is 96, achieved at x = 3.

Common misconceptions to avoid:
- Thinking the optimum could be a maximum for a > 0 quadratic. The positive leading coefficient guarantees a minimum at the vertex, not a maximum. 
- Believing there are multiple local minima or maxima in a simple quadratic; there is exactly one extremum point, the vertex.

In summary, x = 3 is the optimum location, and the optimum type is a minimum (specifically the global minimum for this quadratic).

ECON3200001.1251 Homework Assignment 1

If we optimize the function f(x)=4x2 - 24x + 132, the optimum will be at x = 3 , which is a local minimum .

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