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Chapter 2: Problem 55

Find the absolute maximum and minimum values of each function, if they exist, over the indicated interval. Also indicate the \(x\) -value at which each extremum occurs. When no interval is specified, use the real line, \((-\infty, \infty)\). $$f(x)=x(60-x)$$

Short Answer

Expert verified
The absolute maximum value is 900 at \( x = 30 \), and there is no minimum value.

Step by step solution

01

Understand the Problem

We need to find the absolute maximum and minimum values of the function over the interval \((-\infty, \infty)\). We also need to find the respective \( x \)-values where these extrema occur.
02

Find the Derivative

To find the extrema, first find the derivative of the function. Given \( f(x) = x (60 - x) \), use the product rule or simplify first: \ f(x) = 60x - x^2. Taking the derivative, \ f'(x) = 60 - 2x.
03

Find Critical Points

Set the derivative equal to zero to find critical points: \ 60 - 2x = 0. Solving for \( x \), we get \ x = 30.
04

Determine the Nature of Critical Point

To determine if \( x = 30 \) is a maximum or minimum, check the second derivative: \ f''(x) = -2. Since \( f''(x) < 0 \), it indicates a concave down parabola, so \( x = 30 \) is a maximum.
05

Evaluate Function at Critical Point

Substitute the critical point \( x = 30 \) back into the original function to find the maximum value: \ f(30) = 30(60 - 30) = 900.
06

Analyze the Interval

Since the function is a downward-opening parabola and extends to infinity in both directions, it does not have a minimum value. As \( x \to \pm \infty \), \( f(x) \to -\infty \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

critical points
In calculus, critical points are where the first derivative of a function is zero or undefined. These points help determine where the function might have local maxima or minima. For our function \( f(x) = x(60 - x) \), we first simplify it to \( f(x) = 60x - x^2 \).
Taking the derivative, we get the first derivative \( f'(x) = 60 - 2x \). Setting the derivative equal to zero gives us the critical point:
\[ 60 - 2x = 0 \] Solving for \( x \), we find:
\[ x = 30 \] Hence, \( x = 30 \) is our critical point.
second derivative test
Once we have the critical points, we use the second derivative test to determine if they are maxima, minima, or saddle points. The second derivative of a function helps us understand the concavity. For our function, the second derivative is \( f''(x) = -2 \).
Since \( f''(x) < 0 \), we conclude the function is concave down at \( x = 30 \).
  • If \( f''(x) > 0 \), the function is concave up and the point is a local minimum.
  • If \( f''(x) < 0 \), the function is concave down and the point is a local maximum.
  • If \( f''(x) = 0 \), the test is inconclusive and further analysis is needed.

Therefore, at \( x = 30 \), the function has a local maximum.
concave down parabola
A concave down parabola means that the function opens downward like an upside-down U shape. This is apparent in the function \( f(x) = x(60 - x) \) after evaluating its second derivative. Since \( f''(x) = -2 \), which is less than zero, our parabola is concave down. The highest point on this type of parabola is the vertex, making \( x = 30 \) the maximum point.
This shape intuitively tells us the maximum value of \( f(x) \) occurs at its peak. Evaluating \( f(30) \) gives us:
\[ f(30) = 30(60 - 30) = 900 \]
Therefore, the absolute maximum value is 900.
interval analysis
Interval analysis involves studying the behavior of the function over a specified interval. Here, our interval is the entire real line, \(( -\in \fty, \in \fty \)).
For the function \( f(x) = x(60 - x) \), we analyze its behavior as \( x \) approaches infinity and negative infinity. As \( x \to \pm \in \fty \), \( f(x) \to -\in \fty \).
This means the function keeps decreasing without bound. Hence, there is no minimum value, only an absolute maximum value at \( x = 30 \), where \( f(30) = 900 \).
Understanding interval behavior helps determine where the function reaches its extremes and how it behaves beyond the critical points.

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Most popular questions from this chapter

Cost and tolerance. A painting firm contracts to paint the exterior of a large water tank in the shape of a half-dome (a hemisphere). The radius of the tank is measured to be \(100 \mathrm{ft}\) with a tolerance of 卤6 in. \((\pm 0.5 \mathrm{ft}) .\) (The formula for the surface area of a hemisphere is \(A=2 \pi r^{2} ;\) use 3.14 as an approximation for \(\pi\).) Each can of paint costs \(\$ 30\) and covers \(300 \mathrm{ft}^{2}\) a) Calculate \(d A,\) the approximate difference in the surface area due to the tolerance. b) Assuming the painters cannot bring partial cans of paint to the job, how many extra cans should they bring to cover the extra area they may encounter? c) How much extra should the painters plan to spend on paint to account for the possible extra area?

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