Math Problem Statement

find the x that maximizes (x-30)*(120000-21x+x^2)

Solution

To find the value of xx that maximizes the function f(x)=(x30)(12000021x+x2)f(x) = (x-30)(120000-21x+x^2), we need to follow these steps:

  1. Expand the function:

    f(x)=(x30)(12000021x+x2)f(x) = (x-30)(120000-21x+x^2)

  2. Differentiate the function to find the critical points:

    First, expand the function: f(x)=x(12000021x+x2)30(12000021x+x2)f(x) = x(120000 - 21x + x^2) - 30(120000 - 21x + x^2) =120000x21x2+x330(120000)+30(21x)30(x2)= 120000x - 21x^2 + x^3 - 30(120000) + 30(21x) - 30(x^2) =x321x2+120000x3600000+630x30x2= x^3 - 21x^2 + 120000x - 3600000 + 630x - 30x^2 Combine like terms: f(x)=x351x2+120630x3600000f(x) = x^3 - 51x^2 + 120630x - 3600000

  3. Take the derivative of f(x)f(x):

    f(x)=ddx(x351x2+120630x3600000)f'(x) = \frac{d}{dx}(x^3 - 51x^2 + 120630x - 3600000) f(x)=3x2102x+120630f'(x) = 3x^2 - 102x + 120630

  4. Set the derivative equal to zero to find the critical points:

    3x2102x+120630=03x^2 - 102x + 120630 = 0

  5. Solve the quadratic equation:

    x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=3a = 3, b=102b = -102, and c=120630c = 120630.

    x=(102)±(102)24312063023x = \frac{-(-102) \pm \sqrt{(-102)^2 - 4 \cdot 3 \cdot 120630}}{2 \cdot 3} x=102±1040414475606x = \frac{102 \pm \sqrt{10404 - 1447560}}{6} x=102±14371566x = \frac{102 \pm \sqrt{-1437156}}{6}

Since the discriminant (Δ=b24ac\Delta = b^2 - 4ac) is negative (1437156-1437156), this indicates that there are no real solutions for xx.

Therefore, we need to evaluate the behavior of the function by other means, such as checking the end behavior of the function and its critical values within the domain we are interested in.

  1. Evaluate the function at critical points within the given domain:

    Given that the function involves physical quantities (e.g., 120000 units), consider the practical range for xx.

  2. Analyze the function's behavior at boundaries or other relevant points:

    Check f(x)f(x) at specific points such as x=30x = 30, or any other practical boundaries.

After completing the analysis, we can determine the value of xx that maximizes the function.

Would you like details on this approach or have any specific points you want to delve into?

Here are some relative questions for further exploration:

  1. What is the physical interpretation of xx in this problem?
  2. Are there constraints on xx that we need to consider?
  3. How can we verify the maximum value in a real-world scenario?
  4. What does the end behavior of the function suggest about potential maxima?
  5. Are there alternative methods to find the maximum value?
  6. How would this problem change if there were additional constraints on xx?
  7. Can we use numerical methods to approximate the maximum value of xx?
  8. What are the implications of a negative discriminant in a real-world context?

Tip: Always check the domain and range of a function, especially when dealing with physical or practical problems, as it can affect the validity of mathematical solutions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Critical Points
Derivatives

Formulas

Quadratic equation formula
Derivative formula

Theorems

-

Suitable Grade Level

Advanced High School / College