Math Problem Statement

Find the absolute maximum value of f(x) = 25/x + 4x on the interval [1, 25].

Solution

To find the absolute maximum value of the function f(x)=25x+4xf(x) = \frac{25}{x} + 4x on the interval [1,25][1, 25], we need to follow these steps:

Step 1: Find the Derivative of f(x)f(x)

To determine the critical points, let's differentiate f(x)f(x) with respect to xx:

f(x)=25x+4xf(x) = \frac{25}{x} + 4x

The derivative f(x)f'(x) is: f(x)=25x2+4f'(x) = -\frac{25}{x^2} + 4

Step 2: Set the Derivative to Zero

Now, set f(x)=0f'(x) = 0 to find the critical points:

25x2+4=0-\frac{25}{x^2} + 4 = 0

Rearrange this equation: 25x2=4\frac{25}{x^2} = 4

Multiply both sides by x2x^2: 25=4x225 = 4x^2

Divide by 4: x2=254x^2 = \frac{25}{4}

Take the square root of both sides: x=52orx=52x = \frac{5}{2} \quad \text{or} \quad x = -\frac{5}{2}

Since x=52x = -\frac{5}{2} is not in the interval [1,25][1, 25], we discard it. So, the critical point within the interval is x=52x = \frac{5}{2}.

Step 3: Evaluate f(x)f(x) at the Endpoints and Critical Point

Now, evaluate f(x)f(x) at x=1x = 1, x=25x = 25, and x=52x = \frac{5}{2}.

  1. At x=1x = 1: f(1)=251+4×1=25+4=29f(1) = \frac{25}{1} + 4 \times 1 = 25 + 4 = 29

  2. At x=25x = 25: f(25)=2525+4×25=1+100=101f(25) = \frac{25}{25} + 4 \times 25 = 1 + 100 = 101

  3. At x=52x = \frac{5}{2}: f(52)=2552+4×52=10+10=20f\left(\frac{5}{2}\right) = \frac{25}{\frac{5}{2}} + 4 \times \frac{5}{2} = 10 + 10 = 20

Step 4: Determine the Absolute Maximum

From the values we computed:

  • f(1)=29f(1) = 29
  • f(25)=101f(25) = 101
  • f(52)=20f\left(\frac{5}{2}\right) = 20

The highest value is f(25)=101f(25) = 101. Therefore, the absolute maximum value of f(x)f(x) on the interval [1,25][1, 25] is:

101\boxed{101}

Would you like more details on any specific step or have any questions?

Related Questions

  1. How do we determine if a critical point is a maximum or minimum within an interval?
  2. Can absolute maximum and minimum values occur only at endpoints?
  3. How would the problem change if the interval was not restricted to [1,25][1, 25]?
  4. What role does the derivative play in finding extreme values of a function?
  5. How can the second derivative be used to confirm maximum or minimum points?

Tip

Always check both endpoints and critical points within the interval when finding absolute maximum or minimum values.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Critical Points
Absolute Maximum and Minimum Values

Formulas

Derivative of f(x) with respect to x
Setting derivative equal to zero to find critical points

Theorems

Extreme Value Theorem

Suitable Grade Level

Grades 11-12